LoRA Probe Activations - Gate, Up, and Down Projections

All projections show: input activations ⋅ A matrix (rank-1 LoRA neuron activations)

Layer 0

GATE_PROJ

Top 16 Positive Activations
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a trapez
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when $9 \
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number, write it
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Top 16 Negative Activations
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.003606)^2 �� (0.0036)^2 =
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(0.008)^2 ) �� sqrt(0.000013
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/(π*0.222)) �� sqrt( (8*8.314
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/(π*0.028)) �� sqrt( (8*8.314
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) = 10^-2.5 �� 3.1623e-3
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0.000064) �� sqrt(0.000077
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.\n\nLinear approximation:\n\nsqrt(x + dx) �� sqrt(x) + dx/(2*sqrt(x
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0.000077) �� 0.0087746
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,150 /4975 �� Compute 4975*119
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). sqrt(7.696) �� 2.774, then multiply by
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150 / 5025 �� let's divide 6,060,
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, π * 0.222 �� 3.1416 * 0
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, 168.547 ��168.55, rounded to three
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00007696) �� 0.008774,
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So sqrt(13e-6) �� 0.0036055
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= sqrt(18.72 ) �� 4.327 pF\n\nNow

UP_PROJ

Top 16 Positive Activations
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$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how
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1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here:
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n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
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this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage
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We need to find y such that y ∈ ��_{x ∈ S} sphere(x, d/s
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/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to
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$m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to
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$\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry
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mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
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$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
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$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this
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99 -1). Therefore, the product is ��_{k=1}^{999
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that:\n\n||xi||2 - 2⟨xi, y⟩ + ||y||2 = d
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/(π*0.028)) �� sqrt( (8*8.314
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/2.\n\nRearranged:\n\n2⟨xi, y⟩ = ||xi||2 + ||
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/(π*0.222)) �� sqrt( (8*8.314
Top 16 Negative Activations
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first problem, combining multiple sparse sets into one via tagging. For problem 2, P_bad-angel
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<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these problems one by one. They
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be of this form?\n\nAlternately, let me try to create y. Assume that y is a
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non-promoted average 47.\n\nLet me try to work through that similarly.\n\nFirst, the original
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can impose some structure. Let me fix k and try expressing y in terms of S.\n\nSuppose that
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, \ldots, S_k \) by tagging each element with its respective index. Formally,
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1 bar instead of 1 atm, but regardless, the formula for speed only depends on temperature,
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altitude to EM in triangle GEM. Let me try to visualize this first. \n\nLet me draw the
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4/3024=744÷24=31; 302
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struggling through a number of different avenues, I'll try to formalize the key steps.\n\nGiven S set
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00:**\n - The product alternates modulo 1000. The first two terms
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Now, let's approach this modulo 8 and modulo 125.\n\nFirst, modulo 8
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and after error), passing is at65, regardless of the scaling. However, in the original case
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include x students with original scores >=60 (but <65) and (N/3 -
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0, we consider the product modulo 8 and modulo 125 separately, then combine the results
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), then to determine α_n, the machine can try to generate all possible s of length p(n)

DOWN_PROJ

Top 16 Positive Activations
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that such a point y exists, which is equidistant to all points in S with distance d/s
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constructed by translating an (assumed) existing equidistant set contained 0 and other points. Then
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orthogonality condition comes automatically from the equidistant condition. So that's interesting. Therefore,
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Therefore, if such a y exists which is equidistant to all x in S with distance d/s
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0.\n\nSo actually, if y is equidistant to x and x' with distance d/s
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the distances are equal, it's called an equidistant set. In Hilbert spaces, such sets
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the key difference is S cannot just be any equidistant set, but because the problem tells us that
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the inner product is zero. So, the orthogonality condition comes automatically from the equidistant
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.\n\nBut in our previous equation coming from the orthogonality condition in the original problem, ⟨x
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2/2 for some c, then the orthogonality condition holds.\n\nTherefore, if we can
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0.\n\nSo the point y must be equidistant from all points in S, with distance d
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But look back at the previous equation derived from orthogonale:\n\nFrom above, using orthogon
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orthogonale:\n\nFrom above, using orthogonality condition:\n\n||y||2 - ⟨
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inner product properties that may not be compatible with orthogonality as required. However, in infinite-dimensional
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this might be possible, but S is an equidistant set, which doesn't necessarily contain a linear
Rollout 1
more structure.\n\nWait, if S is an equidistant set, then maybe the function f(x)
Top 16 Negative Activations
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10P5. Yeah.\n\nBut perhaps another perspective: if she's selecting 5 cards from
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x' - y⟩ = 0.\n\nSo another way to think of this is that the set {
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which here is impossible. Therefore, total number of unordered pairs would be \(2^{8 - 1
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'||2 - d2)/2.\n\nBut perhaps another way is expressing this by defining various terms. Let
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65 has a new score70. But another with original60 has new65, but
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^k ordered pairs. Thus, the number of unordered pairs, accounting for swapping a and b, is
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each subset has a unique complement, the number of unordered pairs is indeed 2^{k-1},
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cross-verify.\n\nWait, let’s think of another approach. For instance, imagine that we consider the
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C2)).\n\nBut wait, this might be another approach. Let's see:\n\nIf I take the
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no pairs where a = b. Hence, each unordered pair {a, b} is counted twice in
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info, hard to say.\n\nWait, here's another thought. In gas experiments, diffusion rates, Graham
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240. That's correct.\n\nAlternatively, another way to think about it: Since the order matters
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. But need to specify.\n\nAlternatively, here's another angle: gas thermometric speeds. At 0
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mistake.\n\nWait, alternative approach. Let me use another angle.\n\nGiven after the increase, promoted average=
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2 = d2 / 2.\n\nSimilarly for another x', we get:\n\n||x'||2 -
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propagations are correct.\n\nWait, perhaps there's another way to approach this. Maybe compute the nominal energy

Layer 1

GATE_PROJ

Top 16 Positive Activations
Rollout 6
if and only if it is polynomial-time Turing reducible to a sparse set. So that's the
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able in poly-time, there's no nondeterminism. Since the angel string is computable in poly
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the midpoints of the legs divides the trapezoid into two regions whose areas are in the ratio
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the trapezoid into two smaller trapezoids? Each with height h/2, if
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third is 999 (three 9s), etc., up to a number with 9
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impossible. Hence under any circumstances,given these constraintsthe answer to part(b) is that there is
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angel}} \) does not provide the nondeterminism inherent in \( \textbf{NP} \
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The line at height k creates a smaller trapezoid with bases a and x and height k.
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oly (canonical example given was unary languages including undecidable ones), then L ∈ P_angel is
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. It is known that P/poly contains undecidable languages (as the advice can be uncomput
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the midpoints of the legs divides the trapezoid into two regions with areas in the ratio
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a single string, and find a collision-resistant hash of α_n. Not sure.\n\nWait, think about
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.\n<|im_start|>user\nOne base of a trapezoid is $100$ units longer than
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0 \). The midline of the trapezoid, which connects the midpoints of the legs
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negative inner product. Which contradicts our desired orthogonality.\n\nAlternatively, this tells us that such
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,5), which is (say) left oustide the square, the lines from G to E
Top 16 Negative Activations
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d2) / 2 = 0.\n\nCompare to the above line:\n\nFrom the sum of norms
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75 m/s. Which is correct.\n\nSo comparing that with Rn-222. So
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x'|| squared / 2. ).\n\nThus, matching one of the components:\n\nSo, substituting the
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, between the two lines GE and GM.\n\nBut integrating between x=0 to x=10:\n\n
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30240. That's correct.\n\nAlternatively, another way to think about it: Since the
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168.5 m/s\n\nSo approximately 168.5 m/s. Let
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68.547 m/s.\n\nSo approximately 168.5 m/s. Let
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) ] ] / 10P5\n\nAlternatively, compute numerator and denominator.\n\nFirst, compute denominator
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6%, consistent with the previous result.\n\nTherefore, combining this with the 2*0.00
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Wait, but these are not similar necessarily. \n\nAlternatively, think algebraically. Let me denote the original
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and x=0 to x=10.\n\nAlternatively, the overlapping area might be the area bounded by
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mod1000\n\nAnd so on, alternating between 109 and891 for
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2 gives 0.8%.\n\nSo, when combining these: sqrt( (0.360
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] / (C1 + C2)\n\nThen combine terms:\n\n= (dC1 / C1
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)=167.2 m/s.\n\nSo approximately 167.2 m/s. Another
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but I might need some condition to ensure that.\n\nAlternatively, using the hypothesis that the mutual distances between points

UP_PROJ

Top 16 Positive Activations
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as \boxed{169 \text{ m/s}}.\n\nHowever, depending on source of problem
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\sqrt{\frac{8RT}{\pi M}}\n\]\n\nwhere:\n- \( R \)
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8 * R * T) / (π * M))\n\nR = 8.314 J
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\) is the temperature in Kelvin,\n- \( M \) is the molar mass of radon
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\} \) and a polynomial-time machine \( M \). By the Meyer's theorem equivalence between \(
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i \leq k \). The machine \( M \) with oracle \( S \) constructs the
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.576\n\nDenominator: π * M = π * 0.222 �
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\( \alpha_n \), and the machine \( M \) queries \( S_L \) for the
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,950)=167.2 m/s. So around 167 m/s
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753 ) �� 475 m/s. Which is correct.\n\nSo comparing that with
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624; sqrt ��169 m/s.\n\nBut no one knows. Wait, but
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2) �� 168.5 m/s\n\nSo approximately 168.5
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C1 / C1 + dC2 / C2 - (dC1 + dC2
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47≈168.547 m/s. So approx 168.5
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3) �� 168.6 m/s.\n\nSo my previous result. So the speed
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33,476)=183 m/s. But since the problem asks for mean,
Top 16 Negative Activations
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with countably or uncountably many vertices.\n\nIndeed, in an infinite-dimensional Hilbert space, one
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would be sqrt(2) times the radius.\n\nIndeed, if you take an orthonormal basis in
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there exists a point y common in all spheres.\n\nThus, applying the finite intersection property:\n\n1. For
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this checks out.\n\nSo all steps are correct.\n\nThus, conclusion: at 298 K,
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impossible averages, leading no N satisfies the given.\n\nThus:\n\n(a) 12, 24
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9 mod125. So that matches.\n\nThus, if we follow the steps, the result should
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fine. Good.\n\nNow modulus125:\n\nAgain, Product is(10^k -1
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to 4, there are two color sequences.\n\nThus, the number of color sequences in Case 2
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speed only depends on temperature and molar mass.\n\nThus, if taking STP as 0°C,
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since a and b must be coprime.\n\nThus, since N=20! has prime factors
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formula involves relative errors as decimals, not percentages.\n\nThus, 0.003606
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the area overlapping is inside square and inside triangle.\n\nThus, the overlapping area is the pentagon formed by
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polynomial, but the set must be sparse.\n\nLet's think. For each n, let’s denote a
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(x) = k for all x ∈ S.\n\nThus, the problem reduces to finding y such that the
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r. But they can be in either order.\n\nThus, for each r, the number of ordered color
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that's exponential.\n\nThis seems not helpful. Let's reorient.\n\nPerhaps apply the result from Problem

DOWN_PROJ

Top 16 Positive Activations
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2.6% total spread. But since normally uncertainty is defined as ± half the total spread,
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sphere centered at y. Alternatively, without loss of generality, perhaps we can assume y is at the
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.\n<|im_start|>user\nIn a mathematics test number of participants is $N < 40$ .
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are\nthe following: \nThe average of all participants is $66$ , that of the
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user\nCompute the mean molecular speed v in the heavy gas radon (Rn) in m/s
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are increased by 5, it might cause some participants who previously failed to now pass. Therefore, the
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65. Originally, the average score of all participants is 66. The average of those who
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We are given:\n\n1. The average of all participants is 66. Therefore, the total score
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the number of repeaters as R. Since all participants are either promoted or repeaters, N = P
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Now, after the 5-point increase, every participant's score is increased by 5. Therefore,
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5. Therefore, the new total score for promoted participants would be 71P + 5P
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can't check, but imagine that standard problems involving heavy gases like radon typically compute the speed under standard
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-point increase. Let’s denote the number of promoted participants as P and the number of repeaters as R
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5. At this point the average of the promoted participants becomes 75 and that of the non-prom
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. At this point\nthe average of the promoted participants becomes $75$ and that of
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to now pass. Therefore, the set of promoted participants might change after the score increase. Hmm, so
Top 16 Negative Activations
Rollout 3
in some contexts, STP is 1 bar instead of 1 atm, but regardless, the formula
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1. But problem states75 and59 instead, which is inconsistent. Hence, my previous assumption
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Alternatively, if we take 273 K instead of 298 K, let's compute
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6! example. For 6!, which had instead three distinct prime factors (2,3,5
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maybe the problem uses R=8.31 instead of 8.314. Let me
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0, and check over a smaller case. Suppose instead that Kathy had 1 red and 1 green
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Let me check with a smaller factorial.\n\nSuppose instead of 20!, take a simpler case.
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Wait, but log n is variable length. If instead, we represent n with O(log n) bits
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the mutual distances between points are all d, this might impose that the spheres intersect.\n\nPerhaps we can use
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.88% when rounded. Alternatively, it might be presented as 0.877%,
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p(n) strings for each n, but this might not be sparse if p(n) is a polynomial
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If all scores are increased by 5, it might cause some participants who previously failed to now pass.
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check with smaller products. For verification.\n\nSuppose instead of going up to 999 9
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So if S contains a linear subspace, this might be possible, but S is an equidistant
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and runs in time poly(|x|). If instead, we have an oracle S_L that contains the
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V have certain number of significant figures. The question might not require rounding, but to check if my initial

Layer 2

GATE_PROJ

Top 16 Positive Activations
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a is (b*(a + b) - a*b)/(a + b)^2) = (
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=10, Δb=15, a+b=5000\n\nCompute b2
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)^2) = (b(a + b - a)) / (a + b)^2 = (
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||x - y|| / sqrt{1 - a^, but this is stretching.\n\nBut since in
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720, gcd(a,b)=1, a < b. Then 720 prime factor
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y = sum_{i=1}^n a_i x_i, looking for coefficients a_i such
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a, b)\) and \((b, a)\) are distinct. To find pairs where \(
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all possible values ​​of $N$ in the case where, after the increase, the average
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corresponds to a < b (because 0 < a/b < 1 implies a < b). Therefore
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{a}{b}\) where \(0 < a < b\) and \(a\) and \(b
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b = 20!\), \(0 < a < b\), and \(\gcd(a,
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gcd(a, b)=1, 0 < a < b, and a * b = 2
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b > a would be 20! / a. Therefore, the number of pairs with a <
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Assign 5 to a, others to b: a=5, b= 144 (
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numbered R1, R2, R3, R4, R5), and same with green (
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_p}, and b = 20! / a.\n\nBut in order to count the pairs where a
Top 16 Negative Activations
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25\n\nMultiply through by y to eliminate the denominator:\n\ny(y + 100) -
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298\)\n - Compute the denominator: \(\pi \times 0.2
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, the mol unit cancels, then kg in denominator; thus overall (kg·m2)/(s
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0.88% (since the first non-zero digit is the 8 in the tenths place
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20.576.\n\nDivide by denominator π * 0.222: π
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can see this as a functional equation. Let's denote f(x) = 2⟨x, y
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- 2d2)/4.\n\nLet me denote C = (||t||2 - 2
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30240= Let's check by dividing numerator and denominator by 240:
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/(π*M))\n\nLet's calculate the numerator and denominator step by step.\n\n8*R*T = 8
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7.7528\n\nDivide by denominator.pi*M=3.1416*
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F. Wait, but when we did the partial derivatives, we got ~4.327 p
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\n\nFor any two spheres, there is a non-empty intersection because the distance squared between centers equals d2
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this inner product to be a constant. Let's denote that constant by α. Then, y must be
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} ) = sqrt(d2) = d.\n\nTherefore, any infinite-dimensional Hilbert space can indeed accommodate
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+54N/4.\n\nConvert to same denominator:\n\n62N/12 +54
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dimensions would intersection of such spheres potentially have a non-empty intersection. In infinite dimensions, balls are not compact

UP_PROJ

Top 16 Positive Activations
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We need to solve for x in:\n\nx �� 109 mod125\n\nx
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��5 mod8.\n\nTherefore, P �� 5 mod8.\n\nAlright, now onto modulo
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P mod 8.\n\nSince 10 �� 2 mod 8, so 10
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9 mod125.\n\nTherefore, P ��109 mod125.\n\nSo summary
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109 mod125\n\nx ��5 mod8\n\nLet x =125
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8 are coprime, m must be ��0 mod8. Therefore, m=8n
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find when (2^k -1) �� 0 mod 8, which would make the
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1*3*(-1) = -3 ��5 mod8.\n\nTherefore, P ��
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, so (2^k -1) �� -1 mod8.\n\nTherefore, for k=
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, (10^k -1) �� -1 mod125 for k≥3
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�5 mod8, hence 5m ��0 mod8, leading to m≡0 mod
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for k ≥3, 2^k �� 0 mod8, so (2^k
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mod8, 5m + 5 ��5 mod8, hence 5m ��
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5. Therefore, 10^3 ��0 mod125, and so for all
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Wait, 2^3 = 8 �� 0 mod 8, and 2^
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-angel �� P. Conversely, P �� P_bad-angel because we can ignore the angel
Top 16 Negative Activations
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56*(N/3)-42*( (N/3) -x).\n\nSimplify:\n\n
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since 8 is even). However, in many applications, rounded up. There's conflict here. But
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)) and the side at x=10 (from (10,0) to (10
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/3) + x ) +59*( (N/3) - x ) =71
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00^2) )^2 + ( (15*2000^2)/(
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the 2*0.004 (from the voltage), gives for energy:\n\nsqrt( (
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= ��[0 to10] ( (10/h x) + (10 -
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0.028)) �� sqrt( (8*8.314*29
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is:\n\n∫[0 to10] [ (5/h x +10 -50/h
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= ��[0 to10] [ (5/h x +10 -50/h
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)*(5−r)! = 5!/( (5 - r)! ) * 5!/(
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= ��[0 to10] [ (1/5 x + 8 ) - (
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0.222)) �� sqrt( (8*8.314*29
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2 ) * Δa )^2 + ( (a2 / (a + b)^2 )
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of non-promoted after:47= [ (B(R -x) ) +5(R -
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5 of the cards in a row in a random order."\n\nWait, the problem doesn't specify whether the

DOWN_PROJ

Top 16 Positive Activations
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b) the meaning of passmark might have been misconsidered.\n\nWait, pass mark is fixed at
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<n, i, b>, where the i-th bit of α_n is b. But in order
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the triangle must be mostly inside the square but partially overlapping, with some part outside? Wait, but if
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't compatible (i.e., containing 0 and overlapping points not constructed from orthogonal vectors).\n\nBut in reality
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(10,10). Then, the overlapping region is the part of this triangle that's inside
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0), (10,10). The overlapping area between this triangle and the square requires calculating the
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square area is 100. So the overlapping area is 80. Therefore, the triangle
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.\n\nBut this is the overlapping area. Because The overlapping area is inside the square and inside the triangle,
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) and apex G outside the square. Then the overlapping area between the triangle and square is a trape
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0,10 -50/h). The overlapping area between the triangle and the square would then be
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50/h). Then connecting these points, the overlapping shape would be a trapezoid with vertices
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=0 and x=10.\n\nBut the overlapping region is the part of the triangle GE to GM
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(10,10). So the area overlapping is inside square and inside triangle.\n\nThus, the
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=10.\n\nSo in other words, the overlapping region is between x=0 and x=1
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so triangle is within. But then how come the overlapping area is 80 then? Because that that
Rollout 9
Therefore verifying that for h=25, the overlapping area is indeed 80. Thus, the
Top 16 Negative Activations
Rollout 8
only concerned with the colors. That is, the actual arrangement of colors in the sequence. So if in
Rollout 6
bin(n) concatenated with α_n, then the actual bits of α_n can be encoded in the oracle
Rollout 8
and green, as well as the permutations of the actual cards.\n\nBut maybe in terms of generating the happy
Rollout 3
but I might have to verify which one is the actual mean speed. Let me think.\n\nMean speed,
Rollout 7
sense. If the scores are increased, then the actual marks that count are the new ones. But the
Rollout 2
000. But from the problem, rather than factorizing each term, might use modulus10
Rollout 3
. In gas experiments, diffusion rates, Graham's law, typically at STP. But the molar
Rollout 8
arrangements with two blocks".\n\nAlternatively, in combinatorics, we can model it for a mixed color layout
Rollout 9
small (h<10), the triangle is entirely within the square, and area would be 5
Rollout 9
on the value of h, the triangle might be entirely inside the square (if h ≤10)
Rollout 3
0} m/s\n\nWait no, perhaps with precise calculation using more decimal places.\n\nLet me recalculate
Rollout 4
0.8775%\n\nBut maybe doing precise sqrt(0.000077
Rollout 9
the square. So triangle GEM would then be entirely within the square. Thus, the area in common
Rollout 0
0! is not a square, it's a bit ambiguous. For each a < b, the corresponding
Rollout 0
this, perhaps think again. The question is a bit more precise: it says that the original rational number
Rollout 1
2) = d.\n\nTherefore, yeah, the setup is kind of like S is a system of points

Layer 3

GATE_PROJ

Top 16 Positive Activations
Rollout 0
is \(\boxed{128}\).<|im_end|>
Rollout 1
y \text{ exists as described}}\n\]<|im_end|>
Rollout 9
\) is \(\boxed{25}\).<|im_end|>
Rollout 5
[\n\boxed{181}\n\]<|im_end|>
Rollout 6
) to decide \( x \in L \).<|im_end|>
Rollout 2
is \(\boxed{109}\).<|im_end|>
Rollout 8
[\n\boxed{157}\n\]<|im_end|>
Rollout 4
{0.88\%}\n\]<|im_end|>
Rollout 4
boxed{0.88\%}.\n\n<|im_start|>answer\nTo find the percentage error in the calculation
Rollout 3
\, \text{m/s}\n\]<|im_end|>
Rollout 1
point } y \text{ exists as described}}\n\n<|im_start|>answer\nGiven an infinite-dimensional Hilbert space \(
Rollout 3
\boxed{169} m/s.\n\n<|im_start|>answer\nTo compute the mean molecular speed \( v
Rollout 5
$x^2/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see
Rollout 5
Final Answer**\n\boxed{181}\n\n<|im_start|>answer\nLet the shorter base be \( b \
Rollout 8
Final Answer**\n\boxed{157}\n\n<|im_start|>answer\nKathy has 5 red cards and
Rollout 7
36}\).\n\n(b) No solution exists.<|im_end|>
Top 16 Negative Activations
Rollout 1
could be uncountable. Hmm.\n\nBut let's think: Suppose that S is a set where all
Rollout 8
n=126.\n\nWait, but let's confirm. Wait, when you sum Case1 and
Rollout 9
GE and GM) with base EM.\n\nSo let's calculate coordinates.\n\nGiven:\n\nE is (10
Rollout 4
0.87%.\n\nBut wait, let's go step by step for verification once more.\n\nFirst
Rollout 9
+ (10 - y)^2\n\nCanceling h^2:\n\ny^2 = (1
Rollout 4
15 )^2\n\nHmm, let's compute these step by step.\n\nFirst, compute the
Rollout 1
set, perhaps this intersection is possible.\n\nBut let's see the distance between any two centers (which are
Rollout 4
because the voltage is squared. \n\nWait, let's check the formula. If a quantity Q is proportional
Rollout 9
distance from G to EM.\n\nWait, maybe visualizing coordinates would help. Let's place EM as the
Rollout 2
8 is clearly odd.\n\nBut to confirm, let's check with smaller products. For verification.\n\nSuppose
Rollout 8
I want to cross-verify.\n\nWait, let’s think of another approach. For instance, imagine that
Rollout 6
_bad-angel would equal P.\n\nWait, let's formalize that. Let L be in P_bad
Rollout 3
uses 298 K.\n\nAlternatively, let's cross-validate.\n\nIf I check online for "mean
Rollout 4
* ΔC2)^2 ).\n\nFirst, let's compute the partial derivatives. Let me denote C1
Rollout 1
able subset of S? Hmm.\n\nAlternatively, let's fix an arbitrary x0 in S and then express
Rollout 9
isosceles with GE = GM.\n\nCalculating GE distance:\n\nGE is from G (10

UP_PROJ

Top 16 Positive Activations
Rollout 9
common to triangle $GEM$ and square $AIME$ is $80$ square units.
Rollout 7
56N/3\n\nFactor x:\n\nx(A -54) +18N =5
Rollout 4
)/(a + b)^2) = (b(a + b - a)) / (a + b
Rollout 9
common to triangle \(GEM\) and square \(AIME\) is 80 square units. We
Rollout 9
a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides of length $10$
Rollout 7
=56N/3\n\nThus:\n\nx(A -42) +14N=5
Rollout 7
=56N/3\n\nTherefore:\n\nx(A -54) =56N/3
Rollout 7
x=0\n\n→-3P +x(A -79)=0\n\nBut P=2
Rollout 1
x - y⟩ = d2 / 2,\n\nwhich is ⟨x, x⟩ - 2
Rollout 1
||y||2 = (d2)/2,\n\n||x2||2 - 2⟨x
Rollout 9
boxed{25}\n\n<|im_start|>answer\nSquare \(AIME\) has sides of length 10 units
Rollout 1
:\n\n⟨x, t - x⟩ = C,\n\n⟨x', t - x'⟩ = C
Rollout 7
/3)\n\nWhich simplifies similar to part(a):\n\nA*x +42*N/3 -4
Rollout 7
3=2N/3.\n\nTherefore,\n\nx(A -54)=2N/3.\n\nBut
Rollout 1
y||2 = -d2 / 2,\n\nSo:\n\n⟨x, y⟩ = (d
Rollout 7
6*(N/3)\n\nHence,\n\nx(A -42) +42*(N/
Top 16 Negative Activations
Rollout 2
09. But wait, but I feel like in this case, the product of all these numbers,
Rollout 0
\(a = b\). So all coprime pairs with \(a \times b = 2
Rollout 1
of S and construct y accordingly.\n\nBut since non-separable Hilbert spaces have orthonormal bases of
Rollout 1
the quadratic form is affine.\n\nBut this seems non-trivial, but perhaps specific to the set S.\n\n
Rollout 3
given more precisely, maybe it's three sig figs as 168 m/s. However
Rollout 0
alternative approach: suppose we recognize that each coprime pair (a,b) with a*b = N
Rollout 0
but when we write a and b as coprime factors, the way I'm thinking is that the
Rollout 3
M=0.222 (three sig figs). So in calculation:\n\n8 * R *
Rollout 0
Result**:\n - The number of coprime pairs \((a, b)\) with \(
Rollout 1
triple intersection be non-empty?\n\nMaybe. But how in R^n?\n\nBut since H is infinite-dimensional,
Rollout 0
! and a < b. Since each coprime pair (a,b) with a < b corresponds
Rollout 1
y are orthogonal, we can create infinite such orthogonality conditions. We can think of the intersection
Rollout 0
had three prime factors. The number of coprime pairs a < b where a*b=72
Rollout 3
s minimum if you consider M and T three sig figs.\n\nHence, the answer is 1
Rollout 0
neq b\). Hence, total coprime pairs are \(2^{8}\), where each
Rollout 3
s 169.\n\nTherefore, three sig figs justify 169 m/s.\n\nTherefore

DOWN_PROJ

Top 16 Positive Activations
Rollout 5
[\n\boxed{181}\n\]<|im_end|>
Rollout 1
y \text{ exists as described}}\n\]<|im_end|>
Rollout 8
[\n\boxed{157}\n\]<|im_end|>
Rollout 0
is \(\boxed{128}\).<|im_end|>
Rollout 4
{0.88\%}\n\]<|im_end|>
Rollout 7
. Perhaps some students had their original score so that despite unrealistic averages, their scores sum up to needed.
Rollout 9
+ (10 - y)^2\n\nCanceling h^2:\n\ny^2 = (1
Rollout 1
a shift and scaling.\n\nSumming up, then despite the difficulty with coordinate system, using the properties of
Rollout 3
\, \text{m/s}\n\]<|im_end|>
Rollout 2
is \(\boxed{109}\).<|im_end|>
Rollout 9
\) is \(\boxed{25}\).<|im_end|>
Rollout 0
n=20. So 20!.\n\nRegardless of the multiplicities (since in order for a
Rollout 0
! but 7^3 does not.\n\nBut exponent of 7 is 2, which is even
Rollout 8
, they are singleton blocks, but adjacent in the sense of being in a single block each. But actually
Rollout 6
, since the problem asks to show it, not just cite, perhaps construction:\n\nGiven L ∈ P_
Rollout 3
1 bar instead of 1 atm, but regardless, the formula for speed only depends on temperature,
Top 16 Negative Activations
Rollout 1
into equations involving t. Let me see.\n\nDenote t = 2y. Then y = t
Rollout 5
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a trapezoid is
Rollout 0
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number, write it as a
Rollout 1
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$
Rollout 2
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when $9 \times
Rollout 3
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in the heavy
Rollout 4
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values $C
Rollout 6
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $L \
Rollout 7
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants is
Rollout 8
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red cards and
Rollout 9
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides of length
Rollout 4
using the square root of sums of squares)?\n\nIn most experimental physics contexts, the percentage error refers to the
Rollout 1
can see this as a functional equation. Let's denote f(x) = 2⟨x, y
Rollout 1
this inner product to be a constant. Let's denote that constant by α. Then, y must be
Rollout 5
5 is 181.\n\n\[\n\boxed{181}\n\]<|im_end|>
Rollout 1
's S in the question. But given a priori S with mutual distance d, can you find such

Layer 4

GATE_PROJ

Top 16 Positive Activations
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Top 16 Negative Activations
Rollout 1
Compare to the above line:\n\nFrom the sum of norms, we have:\n\n-⟨x + x',
Rollout 9
this first. \n\nLet me draw the square AIME. Since it's a square, all sides are
Rollout 9
of overlap between triangle GEM and the square AIME is 80.\n\nBut how does the triangle
Rollout 1
, if we have a set of vectors with equal norms and their differences have equal norms, this often relates
Rollout 1
, S' was constructed by translating an (assumed) existing equidistant set contained 0 and
Rollout 9
the common area between triangle GEM and square AIME is 80. Since the square is from
Rollout 9
the area common to triangle GEM and square AIME is 80 square units. The square area
Rollout 3
222). However, depending on the isotope. Natural radon is mainly Rn-2
Rollout 5
length x can be found using the formula which is sort of a weighted average. Let me recall that if
Rollout 3
-219. But the most common isotope is Rn-222, which has
Rollout 3
If Rn-222 is the isotope, then its molar mass is 22
Rollout 2
8. Since 5 and8 are coprime, m must be ��0 mod8.
Rollout 0
prime factors. Each distinct assignment gives a coprime pair (a,b) where a and b are
Rollout 3
222 (since the most stable isotope is radon-222). Wait,
Rollout 1
relate this inner product.\n\nBut without assuming the various norms ||x|| are the same, maybe we can
Rollout 9
is called AIME. If moving around square AIME, probably vertices in order A-I-M-E to

UP_PROJ

Top 16 Positive Activations
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 2
1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here:
Rollout 9
$\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry
Rollout 3
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 4
this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage
Rollout 5
/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to
Rollout 8
$m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to
Rollout 0
$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how
Rollout 1
with y can be controlled.\n\nAlternatively, let's think in terms of coordinates. Suppose we select a point
Rollout 6
, but the set must be sparse.\n\nLet's think. For each n, let’s denote a unique
Rollout 9
to EM, h can be anything. Let's think. If h is less than 10,
Rollout 8
want to cross-verify.\n\nWait, let’s think of another approach. For instance, imagine that we
Rollout 6
perhaps the encoding has to be different. Let's think of ways the oracle can recover α_n efficiently.
Rollout 6
one and keep the combined set sparse? Let's think. Suppose each S_i has at most p(n
Top 16 Negative Activations
Rollout 9
y-coordinate) and M (10 in y-coordinate). That makes sense for an isosceles
Rollout 9
5, halfway between E (0 in y-coordinate) and M (10 in y-coordinate).
Rollout 6
\textbf{P}_{\text{bad-angel}} \subseteq \textbf{P}
Rollout 6
\textbf{P}_{\text{bad-angel}} \), unless \( \textbf{
Rollout 3
, which is a bit different from the root mean square speed. But maybe in this context, they just
Rollout 6
\textbf{P}_{\text{bad-angel}} \) does not provide the nondetermin
Rollout 6
P} =\textbf{P}_{bad-angel}$ ? Is $\textbf{NP
Rollout 6
{NP}=\textbf{P}_{bad-angel}$ ? Justify.\n [/*]\n
Rollout 6
\textbf{P}_{\text{bad-angel}} \), but \( \textbf{
Rollout 6
\textbf{P}_{\text{bad-angel}} \).\n\nFor \( \textbf{
Rollout 6
\textbf{P}_{\text{bad-angel}} \) unless \( \textbf{
Rollout 4
let's perhaps take extreme values. But since error propagation can be complex for combination.\n\nFirst, compute C
Rollout 4
problem, it's likely expecting the partial derivatives error propagation.\n\nBut given that the max/min difference is about
Rollout 6
\textbf{P}_{\text{bad-angel}} \) by setting \( A(n)
Rollout 9
=> y = 5.\n\nTherefore, the y-coordinate of point G must be 5, halfway between
Rollout 4
verify to make sure.\n\nThe equivalent capacitance partial derivatives calculation seems correct. The key point is, for

DOWN_PROJ

Top 16 Positive Activations
Rollout 1
/ 2 = 0.\n\nRearranging:\n\n||y||2 - ⟨x + x
Rollout 1
is a constant depending only on t.\n\nLet me rearrange this as:\n\n⟨x, t - x
Rollout 9
,0).\n\nWait, wait. Let's step back. The triangle GEM is outside except for the
Rollout 1
) / 2 = 0.\n\nRearranging:\n\n||y||2 - ⟨x +
Rollout 1
||2 = d2/2.\n\nRearranged:\n\n2⟨xi, y⟩ = ||
Rollout 1
⟩ - d2 / 2.\n\nSo, rearranged:\n\n⟨x, y⟩ = (⟨
Rollout 1
a constant depending only on t.\n\nLet me rearrange this as:\n\n⟨x, t - x⟩
Rollout 1
||x'||2 = d2.\n\nWhich rearranges to:\n\n⟨x, x'⟩ = (
Rollout 9
5/h)(x -10), which rearranges to y = (-5/h)(x -1
Rollout 1
+ ||x'||2 = d2.\n\nWhich rearranges to:\n\n⟨x, x'⟩ =
Rollout 3
K.\n\nCompute v_avg = sqrt(8*R*T/(π*M))\n\nLet's calculate the numerator and
Rollout 3
m/s: when we compute sqrt((8*R*T)/(π*M)), R is in J/(mol
Rollout 1
2 = d2/2.\n\nRearranged:\n\n2⟨xi, y⟩ = ||xi
Rollout 9
(-5/h)(x -10), which rearranges to y = (-5/h)(x -
Rollout 1
y||2 = d2/2.\n\nRearranged:\n\n2⟨xi, y⟩ =
Rollout 1
2 ) / 2 = 0.\n\nRearranging:\n\n||y||2 - ⟨x
Top 16 Negative Activations
Rollout 3
So approx 161 m/s.\n\nSo depending on the temperature assumption, the answer would be different
Rollout 9
y=0 to y=10. So depending on the value of h, the triangle might be
Rollout 7
59$ .\n(a) Find all possible values ​​of $N$ .\n(b) Find all
Rollout 3
161.4 m/s.\n\nSo, depending on the temperature assumed (273 K vs
Rollout 3
weight is about (222). However, depending on the isotope. Natural radon is mainly
Rollout 0
assigned to a must go completely into a) and then b is 20! / a, with
Rollout 3
9 \text{ m/s}}.\n\nHowever, depending on source of problem, maybe requires two sig fig
Rollout 5
trapezoid into two equal areas, and then find the greatest integer not exceeding x2/1
Rollout 1
three spheres is likely a finite set or empty. Depending on the configuration. For regular simplex, in
Rollout 4
.88%, the answer would differ. So depending on the method, the answer is different. The
Rollout 2
109 and891. Therefore, depending on the number of such terms, if it's
Rollout 3
"mean molecular speed v"—maybe the user is referring to the rms speed? Hmm. Wait, in
Rollout 8
block must be of a single color, and there can only be two blocks, so the layout is either
Rollout 6
α_n, M can compute bin(n) and then generate s by appending alpha_n's bits. But
Rollout 6
p is a polynomial, for each m, there can be multiple n such that p(n) = m
Rollout 1
let's fix an arbitrary x0 in S and then express all other x in S relative to x0

Layer 5

GATE_PROJ

Top 16 Positive Activations
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Top 16 Negative Activations
Rollout 6
n + n^k �� m? Unlikely. For example, m is dominated by n^
Rollout 1
vectors could form an orthogonal set with each vector having norm d / sqrt(2}, so that the distance
Rollout 6
, no, because the machine M doesn't have prior knowledge of α_n. The machine M needs to
Rollout 1
must be an orthogonal set, with each vector having norm d / sqrt(2). Then scaling by sqrt
Rollout 0
will satisfy a < b, even for coprime divisors. Wait, but in reality, since
Rollout 1
finite-dimensional space, in order to have equidistant points, they must form a regular simplex, which
Rollout 0
*b=N!:\n\nSince a and b are coprime.\n\nHence equivalence between choices over subsets of
Rollout 1
y is the origin. Then, all x have norm d / sqrt(2)), and are pairwise orthogonal
Rollout 0
a and b? But since they must be coprime, it's either assigned entirely to a or
Rollout 2
10 and 125 are coprime? 10 and 125
Rollout 0
assignments will satisfy a < b, even for coprime divisors. Wait, but in reality,
Rollout 1
that { x - y } is orthogonal system with norm d / sqrt(2).\n\nFrom Rudin or
Rollout 1
the set {x - y} is orthogonal with norms d/sqrt(2).\n\nAlternatively, is this
Rollout 0
. When we say that a and b are coprime with a*b=20!, then a
Rollout 0
in general, 2^{k} ordered coprime pairs (a,b), and because n!
Rollout 1
might be possible, but S is an equidistant set, which doesn't necessarily contain a linear subs

UP_PROJ

Top 16 Positive Activations
Rollout 1
where each pair is separated by d, perhaps S must lie on such a quadratic manifold. But unless I
Rollout 4
0.88%, or maybe the answer is exact at 0.87%.\n\nBut wait
Rollout 6
poly-time machine to access the bits. Another approach might be to store α_n in the oracle using a
Rollout 3
298 K (25°C), answer is approximately 168.5 m/s.\n\n
Rollout 7
=98.\n\nBut that's impossible as A must be <=64 (since original repeaters with
Rollout 1
(2) } is not convex. Oh, right. The sphere is not convex. So perhaps closed
Rollout 3
168.5\n\nThus, answer is approximately 168.5 m/s.
Rollout 2
the remainder is109.\n\nTherefore, answer:109.\n\nBut let me recheck all
Rollout 3
298 K. In which case, answer is approx 169 m/s.\n\nTherefore,
Rollout 0
8, so k=8. so answer is 2^(8-1)=128
Rollout 2
So need to verify.\n\nBut here is perhaps a mistake: in the modulus8 calculation, concluded product≡
Rollout 3
, let's check some resource. Wait, can't access it.\n\nAlternatively, in absence of the temperature
Rollout 1
able I, the support of each x ∈ H must be countable. So actually, S would be
Rollout 6
-time machine to access the bits. Another approach might be to store α_n in the oracle using a binary
Rollout 5
is 181. Therefore, final answer is 181. \n\nI think that's
Rollout 3
on", do I get a number? But can't check, but imagine that standard problems involving heavy gases
Top 16 Negative Activations
Rollout 8
$m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to
Rollout 3
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 2
1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here:
Rollout 0
$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how
Rollout 4
this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage
Rollout 9
0 units. Isosceles triangle \(GEM\) has base \(EM\), and the area
Rollout 9
$\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this
Rollout 9
altitude to \(EM\) in \(\triangle GEM\).\n\n1. **Vertices of the Square**
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 5
/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to
Rollout 9
units. Isosceles triangle $GEM$ has base $EM$ , and the area
Rollout 9
the altitude to $EM$ in $\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay,
Rollout 9
to x=10:\n\nThe area under line GE from x=0 to x=10 is
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 9
2. ** Isosceles Triangle \(GEM\)**:\n - Base \(EM\) is

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
focus might specifically align with one temperature.\n\nGiven all that, but as per my knowledge cutoff is 2
Rollout 1
the set S into such's direction.\n\nConsidering all that, after struggling through a number of different avenues,
Rollout 3
: 222 g/mol (3 sig figs)), R is 8.314
Rollout 3
s 169.\n\nTherefore, three sig figs justify 169 m/s.\n\nTherefore
Rollout 3
olar mass is 222 (3 sig figs), R is 8.314
Rollout 1
finite-dimensional space, in order to have equidistant points, they must form a regular simplex, which
Rollout 3
depending on source of problem, maybe requires two sig figs. 222 is three, so
Rollout 3
is 8.314 (4 sig figs), T is 298 K (
Rollout 3
98). So the final answer with three sig figs: 169 m/s.\n\n Thus
Rollout 3
98 (3), the least number of sig figs is 3, so result should be three
Rollout 3
is 3, so result should be three sig figs. Hence, 168.5
Rollout 3
68.55, rounded to three sig figs 169.\n\nTherefore, three sig
Rollout 3
given more precisely, maybe it's three sig figs as 168 m/s. However
Rollout 3
T is 298 K (3 sig figs assuming given as 298). So
Rollout 1
distances are equal, it's called an equidistant set. In Hilbert spaces, such sets have
Rollout 1
by translating an (assumed) existing equidistant set contained 0 and other points. Then,
Top 16 Negative Activations
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 6
with a sparse oracle, which is standard.\n\n<|im_start|>answer\n**Problem 1:**\n\nTo construct the
Rollout 3
boxed{169} m/s.\n\n<|im_start|>answer\nTo compute the mean molecular speed \( v \
Rollout 4
{0.88\%}.\n\n<|im_start|>answer\nTo find the percentage error in the calculation of
Rollout 0
Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven a rational number between 0 and
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>

Layer 6

GATE_PROJ

Top 16 Positive Activations
Rollout 8
)*(5−r)! = 5!/( (5 - r)! ) * 5!/(
Rollout 1
we have such a y, then the vectors { sqrt(2)/d(x - y) : x
Rollout 1
I need to find a point y such that { sqrt(2)/d (x - y ) }
Rollout 4
for the uncertainty ΔC_eq would be sqrt( (��C_eq/∂C1 * ΔC
Rollout 9
1/5 x + 8 ) - ( (-1/5 x + 2) ) ]
Rollout 9
+10 -50/h ) - ( (-5/h x) +50/h ) ]
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Top 16 Negative Activations
Rollout 8
sequences: 4P2=12\n\nHappy sequences: All red (needs 2 reds
Rollout 9
$\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry
Rollout 1
onality condition equality. But then, this seems circular.\n\nAlternatively, setting t = ||x|| squared
Rollout 7
total for original promoted is76*P.\n\nNew promoted (original repeaters with60-6
Rollout 4
this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage
Rollout 3
161.4 m/s.\n\nAs an assistant, I need to perhaps suggest both possibilities, but
Rollout 1
equilateral triangle with side length d. Suppose their positions are x1, x2, x3 forming
Rollout 3
mean (average) speed derived from the Maxwell-Boltzmann distribution:\n\n\[\nv = \sqrt
Rollout 3
.5 m/s.\n\nBut user hasn't specified temperature, but as a standard value, that could vary
Rollout 2
109 mod125. So that matches.\n\nThus, if we follow the steps, the
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 4
0036 as a relative error.\n\nThen Voltage's relative error is 0.02/
Rollout 3
61 and 169, depending on temperature. Since the user doesn't specify, but given
Rollout 3
, so need to make a call. \n\n Since chemistry often uses STP as 273K
Rollout 4
77%.\n\nTherefore, the percentage error in energy stored is approximately 0.88%.\n\nWait,
Rollout 9
0 -20=80. So that works.\n\nSo the altitude from G to EM is h

UP_PROJ

Top 16 Positive Activations
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this
Rollout 3
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 2
1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here:
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 0
$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how
Rollout 8
$m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 4
this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides of
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red cards
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants is
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $L
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values $
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in the
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when $9 \times
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H
Top 16 Negative Activations
Rollout 3
(average) speed derived from the Maxwell-Boltzmann distribution:\n\n\[\nv = \sqrt{\
Rollout 9
0), the altitude to EM would just be the horizontal distance from G to the line EM. Wait,
Rollout 6
for large n, so again, each m corresponds to at most one n. Hence, S_L is
Rollout 0
pair (a,b) with a < b corresponds to the exact opposite pair (b,a) with b
Rollout 1
if and only if the distance between x1 and x2 is ≤ 2r. Here, distance
Rollout 9
height' in the trapezoid is the horizontal distance between the vertical sides, which is 1
Rollout 0
\). Hence, each coprime pair corresponds to a subset of the prime factors of 20
Rollout 9
is calculated by integrating the difference between the upper and lower lines from \(x = 0\) to \(
Rollout 9
since EM is vertical, the altitude would be the horizontal distance from point G to the line x=1
Rollout 9
10.\n\nWhich is y = (5/h)x + (10 - 50/h).\n\n
Rollout 4
58 nJ. The difference between maximum and nominal is +0.1958 nJ
Rollout 1
d sqrt(2) �� 1.414d >= d, so intersection is non
Rollout 3
mean (average) speed derived from the Maxwell-Boltzmann distribution:\n\n\[\nv = \sqrt
Rollout 9
. The altitude to EM is simply h, the horizontal distance from G to EM.\n\nWait, maybe visual
Rollout 9
If the altitude is, as I thought, the horizontal distance from G to EM (the line x=
Rollout 8
per color sequence. Wait, but each k corresponds to two color sequences, so for k=2,

DOWN_PROJ

Top 16 Positive Activations
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 1
set S might be uncountable, but orhonormal systems in Hilbert spaces must be countable
Rollout 0
are 8 distinct primes.\n\n2. **Coprime Pairs**:\n - For \(a
Rollout 7
60) is47.\n\nLet me re-do the analysis for part(b):\n\nOriginal promoted = P
Rollout 1
map the equidistant points into orthonornal through such a shift and scaling.\n\nSumming up
Rollout 1
_x,\n\nwhere { e_x } is an orthonormal system.\n\nTherefore, the given set S is
Rollout 0
, 1 and 720 were co-prime (a=1, b=72
Top 16 Negative Activations
Rollout 2
9 \times \cdots \times \underbrace{99\cdots9}_{\text
Rollout 2
9 \times \cdots \times \underbrace{99\cdots9}_{\text
Rollout 0
, but in reality, since a is a number composed by assigning primes to a (possibly a = product
Rollout 3
's choice. Given a chemistry book, Oxtoby, example problem: At 25°C,
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 4
0.88% error, whereas the worst-case scenario would lead to higher.\n\nWait, compute both
Rollout 3
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 0
$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how
Rollout 2
1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here:
Rollout 1
0.\n\nSo the point y must be equidistant from all points in S, with distance d
Rollout 4
the same direction, which is more of a worst-case scenario. Which does the question expect? The question
Rollout 1
- The set \( S \) is equidistant, with all pairs of points distance \( d
Rollout 2
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when $9 \times
Rollout 0
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number, write it as a

Layer 7

GATE_PROJ

Top 16 Positive Activations
Rollout 4
0.88%, or maybe the answer is exact at 0.87%.\n\nBut wait,
Rollout 3
169: 1.685e2 is 168.5, rounded
Rollout 3
exact)( since formula has 8, which is exact), R=8.314,
Rollout 4
00077 is 7.7e-5. sqrt(7.7e-
Rollout 3
answer expects 170 m/s as an approximate answer. But some precision is required, so
Rollout 3
168.5\n\nThus, answer is approximately 168.5 m/s. Thus
Rollout 3
169.\n\nTherefore, three sig figs justify 169 m/s.\n\nTherefore the valid
Rollout 4
) )^2 )\n\nWait, this is another way of writing the terms. Maybe this will increase precision
Rollout 0
and 144. How is it multiplicative, related to the number of prime factors?\n\nYes
Rollout 3
says "mean".\n\nAlternatively, maybe the problem conflates the two, but the term mean is usually average
Rollout 3
precise calculation using more decimal places.\n\nLet me recalculate the exact value.\n\n8 * 8.3
Rollout 4
Given all the uncertainties are in two significant figures at most, perhaps the answer should be to two significant figures
Rollout 7
.\n\nTherefore, this is impossible. Hence under any circumstances,given these constraintsthe answer to part(b
Rollout 7
no solution. Hence possible answer is NONE. But possibly N=12 is hidden but impossible via contradictory
Rollout 3
with one temperature.\n\nGiven all that, but as per my knowledge cutoff is 2023,
Rollout 3
However, based on the inputs: 8 (exact)( since formula has 8, which is exact
Top 16 Negative Activations
Rollout 6
with a sparse oracle, which is standard.\n\n<|im_start|>answer\n**Problem 1:**\n\nTo construct the
Rollout 1
distance between any two distinct points in $ S$ is equal to $ d$ . Show
Rollout 1
} y \text{ exists as described}}\n\n<|im_start|>answer\nGiven an infinite-dimensional Hilbert space \( \
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 7
36}\n\n(b) No solution exists.\n\n<|im_start|>answer\nGiven a mathematics test with \( N \)
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there

UP_PROJ

Top 16 Positive Activations
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 5
125(y + 100)]/y) = 125\n\nMultiply through
Rollout 8
* (5! /k! ) ] ] / 10P5\n\nAlternatively, compute numerator
Rollout 3
us to state that the temperature is required? But no, the problem says "compute the mean molecular speed
Rollout 3
radon (Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to
Rollout 4
C_eq = (C1 * C2) / (C1 + C2). Then, the
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME
Top 16 Negative Activations
Rollout 3
28,392.25\n\nDifference between 28,408.1
Rollout 8
−r) *(5−r)! (choosing 5−r greens and permuting them
Rollout 1
2:\n\n-⟨x + x', y} { x + x', y} }⟩ +
Rollout 8
10P5. If they are indistinct, then the number of distinct color sequences is
Rollout 8
5 green cards, with each color being indistinct, then the number of distinct color sequences is
Rollout 6
? Well, NP is different. In the angel model, you have this poly-length string that can depend
Rollout 1
orthonormal system. Thus, if we can identify y as the center of such a system. The
Rollout 7
, all scores are increased by5. At this point the average of the promoted participants becomes75 and
Rollout 8
(5, r) * r! (choosing r reds from 5 and permuting
Rollout 1
-empty.\n\nGenerally, the intersection of two spheres of radii r1 and r2 with center distance l
Rollout 1
helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$ be an infinite-dimensional
Rollout 2
9 \times \cdots \times \underbrace{99\cdots9}_{\text
Rollout 1
by ⟨x - y, x’ - y}, it is possible? But this seems too vague.\n\n
Rollout 0
would give the count.\n\nBut since in each such coprime pair of divisors, (a,b
Rollout 7
59(R - x).\n\nAdditionally, we can relate total scores based on original averages.\n\nOriginal total score
Rollout 1
S} sphere(x, d/sqrt(2}}.\n\nWe can think of each equation ||x -

DOWN_PROJ

Top 16 Positive Activations
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 6
. The challenge is to encode all the S_i's into S so that, given x and i,
Rollout 6
). If S is the union of all S_i's, but tagged with their index i. Wait,
Rollout 1
any two points x, x' in S are such that x - y and x' - y are
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this
Rollout 6
example, all unary languages, even $UHALT$ which is undecidable, are
Top 16 Negative Activations
Rollout 3
\boxed{169 \text{ m/s}}.\n\nHowever, depending on source of problem,
Rollout 3
three sig figs justify 169 m/s.\n\nTherefore the valid boxed answer is \boxed{
Rollout 3
\(169 \, \text{m/s}\).\n\n**Final Answer:**\n\[\n\
Rollout 3
68.5 \, \text{m/s}\n\]\n\n4. **Rounding to Significant
Rollout 3
0)= approx 161.4 m/s.\n\nAlternatively, precise sqrt(2605
Rollout 3
169} \, \text{m/s}\n\]<|im_end|>
Rollout 3
)= approximately 161.35 m/s.\n\nTherefore, approximately 161.4
Rollout 3
three sig figs is 169 m/s. Thus, present answer boxed as \boxed{
Rollout 3
three sig figs: 169 m/s.\n\n Thus, the mean molecular speed of radon
Rollout 3
answer around \boxed{169} m/s. But need to specify.\n\nAlternatively, here's
Rollout 3
) �� 168.5 m/s\n\nSo approximately 168.5 m
Rollout 3
) is approximately 168.6 m/s. So about 169 m/s.
Rollout 3
Therefore, approximately 161.4 m/s.\n\nSo, depending on the temperature assumed (2
Rollout 3
STP), 161.4 m/s.\n\nAs an assistant, I need to perhaps suggest
Rollout 3
nearest meter per second, 169 m/s. Alternatively, keeping one decimal, 16
Rollout 3
0)= approximately 167.2 m/s. So ~167 m/s.\n\nBut

Layer 8

GATE_PROJ

Top 16 Positive Activations
Rollout 3
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 4
this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 6
with a sparse oracle, which is standard.\n\n<|im_start|>answer\n**Problem 1:**\n\nTo construct the
Rollout 0
$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how
Top 16 Negative Activations
Rollout 8
a random order. She will be happy if and only if all the red cards laid out are adjacent and
Rollout 9
. Since point G is the apex of the triangle opposite the base EM, which is the vertical side of
Rollout 9
0/h <5, because h>10 => 50/h <5. Similarly,
Rollout 5
= 2(b + 75) \implies 3b + 75 =
Rollout 9
So 20y = 100 => y = 5.\n\nTherefore, the y-coordinate
Rollout 5
the legs, has a length equal to the average of the two bases, \( b + 50
Rollout 5
the legs) has a length equal to the average of the two bases. Also, if there is a
Rollout 7
\n\nSo 15P=10N => 3P=2N => P = (
Rollout 5
The segment that joins the midpoints of the legs divides the trapezoid into two regions whose areas
Rollout 0
frac{a}{b}\) where \(0 < a < b\) and \(a\) and \(
Rollout 0
times b = 20!\), \(0 < a < b\), and \(\gcd(a
Rollout 2
9 \equiv 5 \mod 8 \implies 5m + 5 \equiv
Rollout 8
laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRR
Rollout 0
8 assignments:\n\n1. Assign none to a => a=1, b=720 (
Rollout 9
EM when h=10 would be (base * height)/2 = (10 * 1
Rollout 8
are adjacent. So, possible happy sequences are like RRGGG, GGGGR, RRRRR

UP_PROJ

Top 16 Positive Activations
Rollout 3
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 4
this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 9
$\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry
Rollout 8
$m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this
Rollout 2
1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here:
Rollout 0
$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how
Rollout 5
/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to
Rollout 4
are independent and random, and combines them in quadrature, while the max-min approach assumes that all errors
Rollout 1
each x in S are consistent.\n\nAnother thought: think of 2y as a functional. Let me
Rollout 4
errors are independent and random, and combines them in quadrature, while the max-min approach assumes that all
Rollout 1
each x in S must lie on the hyper-quadric defined by ⟨x, t - x⟩
Rollout 3
(average) speed derived from the Maxwell-Boltzmann distribution:\n\n\[\nv = \sqrt{\
Rollout 4
2))) ΔC1 ]^2 \n\nMultiply numerator and denominator:\n\n= [ (C2 ΔC
Top 16 Negative Activations
Rollout 8
240=31/126≈0.246.\n\nThus, the answer
Rollout 0
a,b) where a < b. So the count is half the number of coprime ordered pairs
Rollout 3
6/0.697433≈27,950. sqrt(2
Rollout 3
68.5 +0.047≈168.547. Hence,
Rollout 2
\equiv 5 \mod 8 \implies 5m \equiv 0 \mod
Rollout 2
\equiv 0 \mod 8 \implies m \equiv 0 \mod 8\n
Rollout 2
remainder is 0. However, given that our calculation says 109, which is not zero
Rollout 2
25. So that matches.\n\nThus, if we follow the steps, the result should be10
Rollout 7
+21)/6=49/6≈8.1667\n\nThus:\n\nA
Rollout 0
where a < b is 4 (by explicit count above). But 2^{3-1}=
Rollout 0
in the count. So how do we reconcile the count.\n\nIf the number of coprime assignments (
Rollout 3
76 / 0.697 �� 28,436. Then take
Rollout 3
6 /0.697433≈26,034.21.\n\n
Rollout 5
00$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve this
Rollout 0
rime pairs (excluding duplicates) would give the count.\n\nBut since in each such coprime pair
Rollout 3
4 /0.697433≈28,406, then sqrt≈

DOWN_PROJ

Top 16 Positive Activations
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 3
Let me make sure.\n\nAccording to Maxwell-Boltzmann distribution, the mean speed is the average of
Rollout 3
(average) speed derived from the Maxwell-Boltzmann distribution:\n\n\[\nv = \sqrt{\
Rollout 0
. Therefore, for the purpose of our count, pairing where 2 and 3 are assigned to a
Rollout 2
, 10^k-1 for anyk is always even? No. When k=1
Rollout 6
equal to P.\n\nIn our case, the problem wants to show that L ∈ P_angel implies that
Rollout 3
is to message user to specify temperature, but as IA cannot, so need to make a call. \n\n
Top 16 Negative Activations
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a trapezoid
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number, write it as
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when $9 \times
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in the
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values $
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $L
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants is
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red cards
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides of
Rollout 1
in $ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need
Rollout 2
is \(\boxed{109}\).<|im_end|>
Rollout 3
\boxed{169} m/s.\n\n<|im_start|>answer\nTo compute the mean molecular speed \( v
Rollout 6
) to decide \( x \in L \).<|im_end|>
Rollout 0
Final Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven a rational number between 0 and
Rollout 1
point } y \text{ exists as described}}\n\n<|im_start|>answer\nGiven an infinite-dimensional Hilbert space \(

Layer 9

GATE_PROJ

Top 16 Positive Activations
Rollout 7
equations, but led to impossibility. Thus, perhaps the answer is different.\n\nAlternatively, maybe N=
Rollout 8
.\n\nSame for k=3.\n\nWait, so perhaps for each k from 1 to 4,
Rollout 6
. But this is impossible in polynomial time.\n\nSo perhaps the answer is not straightforward, hence why reference to
Rollout 9
has a portion inside and a portion outside. So perhaps when h is smaller than 10? Wait
Rollout 1
their radius is d / sqrt(2}. So perhaps the pairwise distance between sphere centers squared is much larger
Rollout 3
I can't compute an exact value. Therefore, perhaps the problem is in a context where they expect the
Rollout 6
p(n), which is not feasible.\n\nSo, perhaps the encoding has to be different. Let's think
Rollout 7
there are constraints I haven't considered.\n\nWait, perhaps non-integer averages? But average A=9
Rollout 6
is the Meyer, I think, theorem.) So perhaps this is the intended approach. That if a language
Rollout 7
. Maybe my initial steps overlooked something.\n\nWait, perhaps the error in initial steps is P =2N
Rollout 1
, right. The sphere is not convex. So perhaps closed balls would be convex, but spheres aren't
Rollout 3
169 m/s.\n\nBut wait, perhaps my calculator computation precision was a bit too picky
Rollout 1
intersection even if finite intersections are non-empty. So perhaps we need the family to be "centered"
Rollout 3
170} m/s\n\nWait no, perhaps with precise calculation using more decimal places.\n\nLet me
Rollout 9
. So that suggests something's wrong.\n\nWait, perhaps my assumption is incorrect. Let's recast this
Rollout 8
adjacent", regardless of their individual identities. Therefore, perhaps we can model the problem where cards are just red
Top 16 Negative Activations
Rollout 1
Hilbert space, let $ d>0$ , and suppose that $ S$ is
Rollout 1
distance between any two distinct points in $ S$ is equal to $ d$ . Show
Rollout 1
>0$ , and suppose that $ S$ is a set of points (not necessarily count
Rollout 1
$ S$ is equal to $ d$ . Show that there is a point $ y
Rollout 1
user\nLet $ \mathcal{H}$ be an infinite-dimensional Hilbert space, let
Rollout 1
able) in $ \mathcal{H}$ such that the distance between any two distinct points
Rollout 1
$ y\in\mathcal{H}$ such that \n\[ \left\{\frac
Rollout 0
floor(20/49)=2 +0=2\n\nPrimes 11: floor
Rollout 1
points (not necessarily countable) in $ \mathcal{H}$ such that the distance
Rollout 0
). For example, primes between N/2 and N cannot come in pairs. So hence exponents for
Rollout 0
floor(20/25)=4 +0=4\n\nPrime 7: floor(2
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number, write it as
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when $9 \times
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in the
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values $

UP_PROJ

Top 16 Positive Activations
Rollout 1
: Let me define y such that y = sum_{i=1}^n a_i x_i
Rollout 1
need to find y such that y ∈ ��_{x ∈ S} sphere(x, d/sqrt
Rollout 8
card sequences = 2 *5! + Sum_{r=1}^4 [2 * P
Rollout 8
total happy sequences = 240 + Sum_{k=1}^4 [2 * (
Rollout 1
}{(x - y) ||·|| (sqrt(2}/d}{(x' - y
Rollout 0
the set of prime factors, then a = product_{p in S} p^{e_p}, and
Rollout 8
Then, probability is [240 + Sum_{k=1}^4 [2 * (
Rollout 3
170} m/s\n\nWait no, perhaps with precise calculation using more decimal places.\n\nLet me
Rollout 3
3.15 K):\n\nCompute as earlier:\n\nsqrt(8*8.314*2
Rollout 7
5 is now passing. So effectively, original score >=60 would now pass.\n\nTherefore, in problem
Rollout 1
, x' ∈ S, we have || (sqrt(2}/d}{(x - y)
Rollout 1
-empty if and only if r1 + r2 >= l and |r1 - r2| <=
Rollout 1
for each x in S, v_x = (sqrt(2)/d)(x - y). Then
Rollout 3
)=167.2 m/s.\n\nSo approximately 167.2 m/s. Another
Rollout 3
169 m/s.\n\nBut wait, perhaps my calculator computation precision was a bit too picky
Rollout 1
find a point y such that when I take (sqrt(2)/d)(x - y), these
Top 16 Negative Activations
Rollout 6
) to decide \( x \in L \).<|im_end|>
Rollout 0
Final Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven a rational number between 0 and
Rollout 2
is \(\boxed{109}\).<|im_end|>
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a trapezoid
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number, write it as
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when $9 \times
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in the
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values $
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $L
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants is
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red cards
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides of
Rollout 2
Final Answer**\n\boxed{109}\n\n<|im_start|>answer\nTo find the remainder when \(9 \
Rollout 8
Final Answer**\n\boxed{157}\n\n<|im_start|>answer\nKathy has 5 red cards and
Rollout 7
36}\).\n\n(b) No solution exists.<|im_end|>

DOWN_PROJ

Top 16 Positive Activations
Rollout 1
istant set have to be of this form?\n\nAlternately, let me try to create y. Assume that
Rollout 1
2 - ||x2||2)/2.\n\nSimilarly, using the distance between x1 and x2
Rollout 2
; 125≡5 mod8.\n\nSimilarly 109 mod8: 10
Rollout 4
00 +6=1206. Alternatively, 5025 * 6 =
Rollout 7
scores before the 5-point increase. Let’s denote the number of promoted participants as P and the number
Rollout 0
2^{k-1}=128. Alternatively, total number of subsets is 2^8
Rollout 1
distance d/sqrt(2) from y.\n\nSimilarly, the inner product condition: for x ≠ x
Rollout 1
+ ||y||2) / 2.\n\nSimilarly, for the point 0 ∈ S', for
Rollout 1
d2)/2 - ||x||2.\n\nSimilarly, for x', we get:\n\n-2⟨
Rollout 2
-1≡-1 mod125\n\nSimilarly, all k≥3: terms are 1
Rollout 5
+ 25) * h/2. Similarly, the lower trapezoid would be ((
Rollout 2
9≡-1 mod1000\n\nSimilarly,10^k -1= -1
Rollout 3
per second, 169 m/s. Alternatively, keeping one decimal, 168.
Rollout 4
C1 + C2)) ]^2\n\nSimilarly, second term:\n\n[ (C1 ΔC
Rollout 8
all green cards in the layout are adjacent (similarly if there are reds and greens, reds
Rollout 8
R1 R2, R2 R1. Similarly, all green: G1 G2, G
Top 16 Negative Activations
Rollout 7
0). Their new scores are original+5 (<=64). Average should be47. So
Rollout 3
K (unless stated otherwise), perhaps answer around \boxed{169} m/s. But need
Rollout 5
the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n<|im_start|>answer\nLet the
Rollout 8
, 5! card sequences.\n\nIf 1<=r<=4:\n\nThen, the color sequences must
Rollout 8
- For each possible number of red cards (0 <= r <=5):\n\nIf r=0: all
Rollout 7
with original scores <60.\n\nTherefore, x <=N/3, and (N/3 -
Rollout 0
answer is 128, corresponding to \boxed{128}? Or wait, wait
Rollout 8
all green or all red).\n\nBut for 1 <= r <=4: 2 * P(5
Rollout 7
no solutions.\n\n**Final Answer**\n\n(a) \boxed{12}, \boxed{24},
Rollout 8
157.\n\n**Final Answer**\n\boxed{157}\n\n<|im_start|>answer\nKathy
Rollout 9
answer is 25. Therefore box form \boxed{25}\n\nWait, that seems correct.\n\n
Rollout 9
the altitude being h=25.\n\nPerfect.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>
Rollout 7
the scores are original +5. Hence now, >=65 passes, new. So original score +
Rollout 7
8.\n\nBut that's impossible as A must be <=64 (since original repeaters with scores >=
Rollout 4
error in the calculation of the energy stored is \boxed{0.88\%}.\n\n<|im_start|>
Rollout 3
this question is ambiguous. Nevertheless, likely answer \boxed{170} m/s. Alternatively,

Layer 10

GATE_PROJ

Top 16 Positive Activations
Rollout 3
to be careful here. Let me check.\n\nI recall the root mean square (rms) speed formula is
Rollout 5
/100. \n\nFirst, let me recall some trapezoid properties. In a tr
Rollout 1
to use the parallelogram law. Let me recall that in a Hilbert space, given two vectors
Rollout 1
are mutually orthogonal after scaling.\n\nWait, let me recall that if you have a regular simplex in R^n
Rollout 5
quickly. \n\nAlternatively, perhaps it is quicker to recall the formula for the line that divides the trape
Rollout 5
which is sort of a weighted average. Let me recall that if you have a line parallel to the bases
Rollout 1
is zero for distinct x, x'.\n\nLet me recall that in Hilbert spaces, if we have a
Rollout 7
(P +x)\n\nFrom previous steps:\n\nLet's write this out:\n\n76P + (A +
Rollout 3
since I can't do that, I need to recall. In some problems, if the temperature isn't
Rollout 1
two distinct vectors are orthogonal.\n\nFirst, let me recall that an orthonormal system requires two things:
Rollout 5
that creates a certain area ratio. Let me also recall that the area of a trapezoid is
Rollout 9
Wait, perhaps my assumption is incorrect. Let's recast this. I need to check whether when h
Rollout 0
would be 128.\n\nBut before closing, verifying with our 6! example was key
Rollout 3
standard temperature.\n\nAlternatively, maybe they expect us to recall the molar mass of radon and the temperature
Rollout 8
5 are green.\n\nWait, perhaps it's better to approach the problem by considering the number of color
Rollout 3
with precise calculation using more decimal places.\n\nLet me recalculate the exact value.\n\n8 * 8.
Top 16 Negative Activations
Rollout 1
= x0 + z' is ||z - z'|| = d. Now, we need to find
Rollout 1
H}$ such that the distance between any two distinct points in $ S$ is equal to
Rollout 1
S' is ||z|| = ||x - x0||, but in the original set S,
Rollout 1
x'||2 = 0 ⇒ ⟨x - x', t⟩ = ||x||2 - ||
Rollout 1
, x' ∈ S, then ⟨x - x', y⟩ = ⟨x, y⟩ -
Rollout 1
sqrt(2),\n\nwe also have ||x - x'|| = d = sqrt(2) * ||
Rollout 1
inner product formula:\n\nd2 = ⟨x - x', x - x'⟩ = ||x||
Rollout 1
d / sqrt(2) ≥ ||x - x'||, which is d sqrt(2) ≥
Rollout 1
2.\n\nWhich gives that inner product ⟨x - x', y⟩ = (||x||2 -
Rollout 1
for some constant c. Then, ⟨x - x', y⟩ would be (⟨x, x
Rollout 1
:\n\nFor each z ∈ { z = x - x0 : x ∈ S }, we have ||z
Rollout 1
x0 ∈ S. Let z = x - x0 for x ∈ S. Then, S =
Rollout 1
then from the above equation, Re⟨x - x', y⟩ = 0.\n\nBut in general
Rollout 1
||x||2 = c and ||x - x'|| = d).\n\nAlso, ⟨x, y
Rollout 1
and x' in S. Then ||x - x'|| = d. Let me assume that there exists
Rollout 1
x'||2 = d2 ⇒ ⟨x - x', x - x'⟩ = d2 ⇒

UP_PROJ

Top 16 Positive Activations
Rollout 1
)/2.\n\nPutting all this together.\n\nNow, suppose that we can model some kind of system.\n\nLet
Rollout 1
+ ⟨y, y⟩ = 0.\n\nWhich is equivalent to:\n\n⟨x, x'⟩
Rollout 1
y||2 = d2 / 2.\n\nWhich can be written as:\n\n-2⟨x,
Rollout 1
/ 2 + d2/ 2.\n\nWhich is equal to:\n\n( d2/2 -
Rollout 1
But this might not make it easier.\n\nAlternatively, suppose that S contains 0, and all other points
Rollout 1
y||2 - (d2)/2. Which previous.\n\nAlternatively, if more generally, for each
Rollout 1
' - y' �� = 0.\n\nWhich brings us back to the original but shifting the problem
Rollout 1
of xi with y is this value.\n\nSo, assuming that all such equations hold.\n\nBut if we fix
Rollout 1
ity. However, let me refocus.\n\nSuppose that I can write each x ∈ S as:\n\n
Rollout 1
}x∈S} is orthonormal.\n\nWhich is equivalent to requiring that { x - y }
Rollout 9
=20 +60=80. Which is correct. Therefore verifying that for h=2
Rollout 4
696)=0.877%.\n\nWhich 0.877% is in percentages
Rollout 7
(N/3 -x)=71N.\n\nWhich simplifies as we did before to x=N/
Rollout 6
can only access poly many bits of α_n. Suppose the original machine M accesses at most q(|x
Rollout 1
by defining various terms. Let me see.\n\nSuppose that all points in S lie in the set {
Rollout 1
⟩ + ||x'||2 = d2.\n\nWhich rearranges to:\n\n⟨x, x'⟩
Top 16 Negative Activations
Rollout 5
zoid into two regions with areas in the ratio 2:3. Then, we need to find
Rollout 9
there's a square named AIME with sides of 10 units. There's also an isos
Rollout 9
-right. So the square is AIME with sides 10 units each. Then we have an is
Rollout 1
two things: 1) each vector has norm 1, and 2) any two distinct vectors
Rollout 1
thonormal system. That means each vector has length 1 and any two distinct vectors are orthogonal.\n\nFirst
Rollout 1
}(x - y) \) must have norm 1. This implies \( \|x - y\
Rollout 9
. Since it's a square, all sides are 10 units. Let's label the square with
Rollout 5
the area into two equal parts, each of area 62.5H.\n\nLet me let x
Rollout 1
in S lie on a sphere with center y and radius d / sqrt(2), then the necessary and
Rollout 5
divides the trapezoid into areas with ratio 2:3, leading us to bases of
Rollout 1
spheres centered at each x in the finite subset with radius d / sqrt(2) must be non-empty
Rollout 9
\nSquare \(AIME\) has sides of length 10 units. Isosceles triangle \(
Rollout 2
99, 999, and so on, up to a number with 99
Rollout 0
numerator and denominator. For how many rational numbers between 0 and 1 will $20_{}
Rollout 1
that all x in S lie on a sphere of radius d/sqrt(2) around y, and
Rollout 2
to find the remainder when the product of numbers like 9, 99, 999

DOWN_PROJ

Top 16 Positive Activations
Rollout 1
the distance between any two is the same, they form a regular simplex. The regular simplex has the property
Rollout 1
But actually no, in the previous, if we take multiple points, we need to ensure that they lie
Rollout 1
Hilbert space \( \mathcal{H} \) and a set \( S \) of points
Rollout 1
non-empty.\n\nSo in our case, if we take any finite subset of S, then the intersection of
Rollout 1
d2 / 2 = d2. Then considering squared distances in the planes, the intersection would be
Rollout 3
answer\nTo compute the mean molecular speed \( v \) of radon (Rn) gas,
Rollout 1
a constant.\n\nBut in that case, if we take x, x' ∈ S, then ⟨x
Rollout 1
certain conditions are satisfied.\n\nWait, for example, take three points in R^3 with equal pairwise distances
Rollout 1
can be considered as an orthonormal basis for some subspace of H. However, since H is
Rollout 5
<|im_start|>answer\nLet the shorter base be \( b \) and the longer base be \( b +
Rollout 1
order to have equidistant points, they must form a regular simplex, which can be embedded in n
Rollout 7
\) participants (\( N < 40 \)), the pass mark is fixed at 65
Rollout 5
longer base be \( b + 100 \). The midline of the trapezoid
Rollout 1
2 = (d2)/2. Now, taking two points x1 and x2 in S,
Rollout 1
2) times the radius.\n\nIndeed, if you take an orthonormal basis in H, and position
Rollout 1
-dimensional, which typically refers to having a countably infinite orthonormal basis, but perhaps the problem allows
Top 16 Negative Activations
Rollout 9
_upper and y_lower from a to b is ��[a to b] (y_upper - y
Rollout 9
+50/h ) ] dx\n\n= ��[0 to10] [5/h x
Rollout 3
16 * 0.222 �� 0.697. So 1
Rollout 9
y_upper - y_lower ) dx\n\n= ��[0 to10] [ (1/
Rollout 3
+ 199.536 �� 18,157.77
Rollout 4
= sqrt(18.72 ) �� 4.327 pF\n\nNow
Rollout 3
76 / 0.697 �� 26,049.\n\nsqrt(
Rollout 3
76 / 0.697 �� 28,436. Then take
Rollout 4
327 / 1200 �� 0.003606.
Rollout 3
0.08798) ) �� sqrt( ~225,753
Rollout 8
total happy sequences = 240 + Sum_{k=1}^4 [2 * (
Rollout 9
GM.\n\nAlternatively, the area is computed by ��[0 to10] ( y_upper -
Rollout 8
the color sequences must be either R^r G^{5−r} or G^{5−r
Rollout 9
+ 2) ) ] dx\n\n= ��[0 to10] [ (1/
Rollout 4
(0.008)^2 ) �� sqrt(0.000013
Rollout 3
6*0.028) ) �� sqrt( (19872.7

Layer 11

GATE_PROJ

Top 16 Positive Activations
Rollout 8
want to cross-verify.\n\nWait, let’s think of another approach. For instance, imagine that we
Rollout 9
to EM, h can be anything. Let's think. If h is less than 10,
Rollout 1
be uncountable. Hmm.\n\nBut let's think: Suppose that S is a set where all pairwise
Rollout 2
25.\n\nThis seems more challenging. Let me think.\n\nWe need to compute P mod125
Rollout 1
the same Hilbert space.\n\nWait, let me think again.\n\nGiven set S wherein between any two points
Rollout 3
which one is the actual mean speed. Let me think.\n\nMean speed, or average speed, is indeed
Rollout 6
perhaps the encoding has to be different. Let's think of ways the oracle can recover α_n efficiently.
Rollout 6
, but the set must be sparse.\n\nLet's think. For each n, let’s denote a unique
Rollout 6
one and keep the combined set sparse? Let's think. Suppose each S_i has at most p(n
Rollout 2
divisible by 1000. Let me think.\n\nBreaking down. Let's check the factors of
Rollout 3
radon gas at a standard temperature. Let me think. If I remember, for example, for gases
Rollout 1
with y can be controlled.\n\nAlternatively, let's think in terms of coordinates. Suppose we select a point
Rollout 1
orthocentric system or something similar?\n\nLet me think.\n\nSuppose we fix y. Then the conditions
Rollout 8
are adjacent and all greens are adjacent.\n\nAlternatively, think of the problem as the laid-out 5 cards
Rollout 1
infinite such orthogonality conditions. We can think of the intersection of all spheres and hyperplanes.\n\n
Rollout 5
but these are not similar necessarily. \n\nAlternatively, think algebraically. Let me denote the original trape
Top 16 Negative Activations
Rollout 1
need to find a point y such that { sqrt(2)/d (x - y ) }x
Rollout 1
e_i} where each e_i is (sqrt(2)/d)(x_i - y) for
Rollout 3
3≈26,050\n\nsqrt(26,050)= approx 1
Rollout 3
3≈27,950\n\nsqrt(27,950)= approximately 1
Rollout 3
,408.123\n\nsqrt(28,408.123
Rollout 3
6,034.21.\n\nsqrt(26,034.21)=
Rollout 4
(from the voltage), gives for energy:\n\nsqrt( (0.0036055
Rollout 1
a point y such that when I take (sqrt(2)/d)(x - y), these vectors
Rollout 0
2 < 20! or a < sqrt(20!). Therefore, the number of such a
Rollout 3
.15 K):\n\nCompute as earlier:\n\nsqrt(8*8.314*27
Rollout 5
[150 ± 10*sqrt(725)] / 2 = 7
Rollout 1
distance each being d and radii d / sqrt(2}) would be non-empty? Not necessarily.
Rollout 4
0.000077\n\nsqrt(0.000077)≈
Rollout 1
(x - y) ||·|| (sqrt(2}/d}{(x' - y}
Rollout 1
x' ∈ S, we have || (sqrt(2}/d}{(x - y) ||
Rollout 4
2*0.4)^2 )=sqrt(0.1296 +0.6

UP_PROJ

Top 16 Positive Activations
Rollout 2
*99*999*... So thinking about modulus 1000, when you
Rollout 9
{25} . Seems done.\n\nThat's logical. Another way:\n\nFrom h=25,
Rollout 8
are adjacent and all greens are adjacent.\n\nAlternatively, think of the problem as the laid-out 5 cards
Rollout 9
.e., to the left of the square. So then, the triangle connects (negative,5), (
Rollout 6
hash of α_n. Not sure.\n\nWait, think about P_angel: the original TM M is
Rollout 1
The problem is over the Hilbert space, so thinking as in R^n, for which we know that
Rollout 0
have a < b or a > b, so then to get the (a,b) with a <
Rollout 3
focus might specifically align with one temperature.\n\nGiven all that, but as per my knowledge cutoff is 2
Rollout 7
via contradictory.\n\nWait maybe but we need more precise thinking. Perhaps some students had their original score so that
Rollout 7
x)\n\nFrom previous steps:\n\nLet's write this out:\n\n76P + (A +5)x
Rollout 0
128?\n\nBut to confirm this, perhaps think again. The question is a bit more precise:
Rollout 8
-color cases which do meet the condition.\n\nTherefore, returning to the original problem, seems that my answer is
Rollout 8
Therefore, for color arrangements where all reds are together and all greens are together, regardless of the order
Rollout 7
) is 75. Wait, that’s not 76. So even though original promoted had
Rollout 9
-100/h ] dx.\n\nYes because combine 5/h x +5/h x =
Rollout 3
8.55, rounded to three sig figs 169.\n\nTherefore, three sig fig
Top 16 Negative Activations
Rollout 9
triangle might be entirely inside the square (if h ≤10) but h is positive.\n\nBut wait
Rollout 0
find all pairs \((a, b)\) such that \(a \times b = 20
Rollout 1
(2) + d / sqrt(2) ≥ ||x - x'||, which is d sqrt
Rollout 6
length m, |S_L^{=m}| ≤ q(m) for some polynomial q.\n\nSo how
Rollout 1
) ≥ d. Which is sqrt(2) ≥ 1, which is true. Therefore, their
Rollout 1
need that for any x, ||v_x|| = 1, and for any x ≠ x',
Rollout 1
in $ \mathcal{H}$ such that the distance between any two distinct points in
Rollout 6
, $|S^{=n}| \leq p(n)$ , where $S^{
Rollout 0
related to the divisors of 20! where each pair consists of two coprime numbers.\n\n
Rollout 1
x'||, which is d sqrt(2) ≥ d. Which is sqrt(2) ≥
Rollout 1
helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$ be an infinite-dimensional
Rollout 0
rime pairs \((a, b)\) where \(a \times b = 20!\
Rollout 1
is an orthonormal system of vectors in $ \mathcal{H}$ .\n<|im_start|>assistant\n
Rollout 1
$ \mathcal{H}$ such that the distance between any two distinct points in $
Rollout 6
| = p(n), and a poly-time M such that x ∈ L iff M(x, α_n
Rollout 6
n, so |S_L^{=m}| ≤1, which is certainly sparse. Thus, S

DOWN_PROJ

Top 16 Positive Activations
Rollout 7
16x =71N -209N/3= (213N -
Rollout 7
x.\n\nTherefore, the equation becomes:\n\n209N/3 +16x =71
Rollout 7
3= (213N -209N)/3=4N/3.\n\nTherefore
Rollout 5
formula:\n\ny = [150 ± sqrt(22500 + 500
Rollout 0
, a can vary from 1 up to sqrt(20!) if possible, but since 2
Rollout 1
a regular simplex in R^n with edge length sqrt(2), then the vectors from the centroid to the
Rollout 7
71N.\n\nNow, subtract 209N/3 from both sides:\n\n16x
Rollout 1
between any two points in the simplex would be sqrt(2) times the radius.\n\nIndeed, if you
Rollout 3
3≈26,050\n\nsqrt(26,050)= approx 1
Rollout 1
a point y such that when I take (sqrt(2)/d)(x - y), these vectors
Rollout 3
,408.123\n\nsqrt(28,408.123
Rollout 1
d2, and their radius is d / sqrt(2}. So perhaps the pairwise distance between sphere centers
Rollout 3
6,034.21.\n\nsqrt(26,034.21)=
Rollout 1
y so that ||y|| = d / sqrt(2), then these vectors would not be orthogonal.\n\n
Rollout 1
our case, each sphere has radius d / sqrt(2} approx 0.707d
Rollout 7
3 +59N/3=209N/3.\n\nSimilarly, the x terms:\n\n
Top 16 Negative Activations
Rollout 4
\pm 0.02 \mathrm{~V}$. What is the percentage error in the
Rollout 4
1200 = let's compute that more accurately.\n\n4.327 / 12
Rollout 5
= 181.25, greatest integer not exceeding 181.25 is
Rollout 4
\approx 0.8775\%\n\]\n\nRounding to two significant figures, the
Rollout 3
is approximately 222 (since the most stable isotope is radon-222).
Rollout 5
into two regions of equal area. Find the greatest integer that does not exceed $x^2/1
Rollout 5
81.25\n\]\nThe greatest integer not exceeding 181.25 is
Rollout 5
181.25, so the greatest integer not exceeding this is 181.\n\nWait
Rollout 7
score increase. Hmm, so that's a critical point.\n\nWait, but reading the problem again: "
Rollout 4
3 �� 8.77e-3, which is 0.0087
Rollout 3
* 0.222)) Hmm, okay. Let me calculate.\n\nFirst, 8 *
Rollout 2
divided by 1000. Hmm, okay, let's unpack this step by step.\n\nFirst
Rollout 7
60-64, leading to a max average of64.\n\nThis is impossible, meaning that
Rollout 9
-right, and E is the bottom-right. So then moving from A to I to M to E and
Rollout 5
181.25, so the greatest integer not exceeding that is 181. Therefore
Rollout 3
2*200=13,302.4 +66.512

Layer 12

GATE_PROJ

Top 16 Positive Activations
Rollout 3
which is 300 - 2, so subtract 66.512 *
Rollout 4
5,025,000, so 5,025,000
Rollout 0
can have a < b or a > b, so then to get the (a,b) with a
Rollout 2
8 (109 mod8=5), so conclusion is x=109 mod10
Rollout 2
91-875=16. So 891≡16 mod12
Rollout 2
9≡-1 mod1000. So 891*(-1)= -89
Rollout 6
) (since p(n) is poly(n), so log p(n) = O(log n)). Therefore
Rollout 2
109, which is5 mod8. So according to Chinese Remainder Theorem, there is
Rollout 8
so 30 +1=31. So 31. Denominator: 30
Rollout 7
5. If scores are increased by 5, so for promotion after the increase, would it be over
Rollout 2
1000= 999. So multiply by999 again. 89
Rollout 2
9≡-1 mod1000. So 109*(-1)=-10
Rollout 0
,17,19 each only once. So actually, their exponents are odd (each exponent
Rollout 2
00, it's 109. So P≡109 mod125 and
Rollout 8
/1! ] = 120. So 5 * 120 = 6
Rollout 2
000, that's109. So 891*999≡1
Top 16 Negative Activations
Rollout 4
's start by recalling how to calculate the equivalent capacitance of capacitors in series and then the energy stored
Rollout 1
as described}}" or as per problem statement.\n\n**[Final Answer]**\n\boxed{\text{Such
Rollout 7
average 47.\n\nLet me try to work through that similarly.\n\nFirst, the original setup: All
Rollout 3
K (unless stated otherwise), perhaps answer around \boxed{169} m/s. But need
Rollout 3
70 m/s, considering rounding, but original exact formula gives 168.5.\n\nBut
Rollout 4
that as well.\n\nThe formula for the equivalent capacitance in series is C_eq = (C1 *
Rollout 9
it's shaped like a quadrilateral?\n\nLet me sketch mentally: triangle GEM has vertex G outside on
Rollout 7
with part (a). Let's split the problem step by step.\n\nFirst, let's consider the original
Rollout 9
, perhaps my assumption is incorrect. Let's recast this. I need to check whether when h increases
Rollout 3
that for 298 K.\n\nV_rms = sqrt(3RT/M) = sqrt(
Rollout 4
let's start by recalling how to calculate the equivalent capacitance of capacitors in series and then the energy
Rollout 4
, for capacitors in series, the equivalent capacitance C_eq is given by 1/C_eq =
Rollout 4
7%.\n\nBut wait, let's go step by step for verification once more.\n\nFirst, Compute C_eq
Rollout 4
}} \) and its uncertainty. The equivalent capacitance for capacitors in series is given by:\n\n\
Rollout 4
let me verify to make sure.\n\nThe equivalent capacitance partial derivatives calculation seems correct. The key point is
Rollout 3
\boxed{169} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon

UP_PROJ

Top 16 Positive Activations
Rollout 3
average speed. But given that different sources can have different conventions, perhaps use both.\n\nBut no, as
Rollout 1
the previous calculation into the orthogonality condition equality. But then, this seems circular.\n\nAlternatively,
Rollout 1
d>0$ , and suppose that $ S$ is a set of points (not necessarily
Rollout 1
answer\nGiven an infinite-dimensional Hilbert space \( \mathcal{H} \) and a set
Rollout 0
Therefore, odd. Hence sqrt(N!) is not integer. Hence N! is not a square. Therefore
Rollout 2
by (-1), etc.\n\nWait, but the third term is999≡-1 mod1
Rollout 6
x|. Therefore, L is in P, so P_bad-angel �� P. Conversely, P
Rollout 6
}} \) has a polynomial-length advice sequence \( \{\alpha_n\} \) and a polynomial
Rollout 3
/s2). Square root is m/s. So unit check.\n\nAnother check: nitrogen gas (28
Rollout 3
, mean speed depends on temperature, so we need T. Since the user hasn't specified temperature, perhaps
Rollout 1
these equations are consistent, that is, that a common y satisfies all of them.\n\nAlternatively, for the
Rollout 1
ormal systems of arbitrary cardinality.\n\nSo, using Zorn's lemma, to find a maximal orthon
Rollout 4
b = 15 pF\n\nSo, a + b = 2000 +
Rollout 6
So unless NP = P, which is unknown, NP would not equal P_bad-angel. But according
Rollout 6
_n computable in poly-time. Is P = P_bad-angel? Is NP = P_bad-
Rollout 6
\( n \), ensuring sparsity as each \( \alpha_n \) is unique per length.\n\n**
Top 16 Negative Activations
Rollout 3
as earlier for radon)\n\nDenominator: pi*0.028= ~0.0
Rollout 3
.15 K):\n\nCompute as earlier:\n\nsqrt(8*8.314*27
Rollout 3
Rn at 298 K is sqrt(3RT/M). Let's compute:\n\n3*
Rollout 1
our case, each sphere has radius d / sqrt(2} approx 0.707d
Rollout 3
3≈27,950. sqrt(27,950)=167
Rollout 3
3≈28,622\n\nsqrt(28,622)= about 1
Rollout 3
m/s. Let's check my math:\n\nsqrt(8*8.314*29
Rollout 4
with coefficients. Then the total energy error is sqrt(0.362 + (2*0
Rollout 4
0.000077. sqrt(0.000077)= approximately
Rollout 3
3≈27,950\n\nsqrt(27,950)= approximately 1
Rollout 1
, ||x - y|| = d / sqrt(2),\n\nwe also have ||x - x'||
Rollout 9
50/h) and (0,10 -50/h)) and the side at x=
Rollout 3
take 298 K first.\n\nCompute sqrt(8 * 8.314 *
Rollout 1
d2, and their radius is d / sqrt(2}. So perhaps the pairwise distance between sphere centers
Rollout 4
V. The relative error in energy would be sqrt( (relative error in C_eq)^2 + (
Rollout 4
V's 0.008:\n\nsqrt(0.00362 +0.

DOWN_PROJ

Top 16 Positive Activations
Rollout 4
\pm 0.02 \mathrm{~V}$. What is the percentage error in the
Rollout 1
\frac{d}{\sqrt{2}} \).\n\n3. ** using Hilbert Space Properties**
Rollout 7
P = \frac{2N}{3} \) and \( R = \frac{N}{
Rollout 1
\frac{d}{\sqrt{2}} \) can be non-empty due to the finite intersection
Rollout 1
\frac{d}{\sqrt{2}} \).\n - For any two distinct \( x,
Rollout 1
helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$ be an infinite-dimensional
Rollout 7
\( R = \frac{N}{3} \), solving gives \( x = \frac{N
Rollout 7
x = \frac{N}{12} \).\n - \( x \) must be integer
Rollout 1
answer\nGiven an infinite-dimensional Hilbert space \( \mathcal{H} \) and a set
Rollout 7
x = \frac{N}{12} \).\n\n2. **Verification leads to contradictions:**\n
Rollout 8
the number of sequences is the number of injective functions from 5 positions to 10 cards,
Rollout 1
Hilbert space \( \mathcal{H} \) and a set \( S \) of points
Rollout 1
+ k.\n\nIf H is a real Hilbert space, then T is a continuous linear functional. Hence
Rollout 5
the two bases, \( b + 50 \). The areas of the regions divided by this mid
Rollout 6
x \) with the binary representation of \( i \). Since each \( S_i \) is sparse
Rollout 5
}}\n\]\nwhere \( a = 75 \) and \( b = 175 \
Top 16 Negative Activations
Rollout 3
\boxed{169} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon
Rollout 4
, 0.360583333... �� 0.00
Rollout 1
exists as described}}" or as per problem statement.\n\n**[Final Answer]**\n\boxed{\text{
Rollout 2
boxed answer is \boxed{109}\n\n**Final Answer**\n\boxed{109}\n\n
Rollout 4
.774 * 3.1623e-3 �� 8.7
Rollout 7
.\n\nTherefore, part(b) has no solutions.\n\n**Final Answer**\n\n(a) \boxed{12
Rollout 7
+5 >=65, that's original score >=60. Wait, but originally, the pass
Rollout 0
,3,5,7 are even?\n\nWait floor operations:\n\nFor prime 2: 20
Rollout 4
004)^2 )\n\nCalculating 2*0.004 = 0.0
Rollout 7
59.\n\nWait, but that suggests that perhaps the increase affects the promotion status? But the problem
Rollout 6
part of the input. Wait, the input is <x, i>, so i is part of the
Rollout 6
more carefully: Let the new set S be { (i, x) | x ∈ S_i },
Rollout 4
= 0.360583333... So approximately 0.36
Rollout 8
+126= 157.\n\n**Final Answer**\n\boxed{157}\n\n
Rollout 4
.7749*3.1623≈8.77.\n\nTherefore,
Rollout 4
2.5 �� 3.1623e-3, so 2.7

Layer 13

GATE_PROJ

Top 16 Positive Activations
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 6
hash of α_n. Not sure.\n\nWait, think about P_angel: the original TM M is
Rollout 9
Wait, hold on. \n\nWait, in the square AIME, let me confirm the positions. If
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 5
/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to
Rollout 8
$m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to
Rollout 9
$\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry
Rollout 5
but these are not similar necessarily. \n\nAlternatively, think algebraically. Let me denote the original trape
Rollout 1
each x in S are consistent.\n\nAnother thought: think of 2y as a functional. Let me
Rollout 2
1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here:
Rollout 1
.\n\nBut how can we solve this? This seems cyclic.\n\nAlternatively, it becomes apparent that:\n\nFor each
Rollout 2
1 mod125.\n\nTherefore, product P mod125 is: 9 *99
Rollout 8
want to cross-verify.\n\nWait, let’s think of another approach. For instance, imagine that we
Rollout 9
, the area of overlap between triangle GEM and square AIME is 80, which is
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this
Rollout 9
10). Wait, but in trapezoid area formula:\n\nAverage of the two bases * height
Top 16 Negative Activations
Rollout 7
constraints missed. Alternatively, perhaps N=24?\n\nLet me check manually for N=24.\n\n
Rollout 3
69. Maybe 168 m/s.\n\nBut checking further on speed, considering the problem might
Rollout 0
, the formula is 2^{k -1).\n\nHence, for 20! with
Rollout 3
is approximately 168.5 m/s.\n\nIn absence of more specifications, but given standard calculations
Rollout 3
Alternatively, perhaps they have in mind 0°C.\n\nAlternatively, maybe the problem wants the root mean square
Rollout 3
decimal, 168.5 m/s.\n\nHowever, considering that we used approximate pi and calculator
Rollout 3
original exact formula gives 168.5.\n\nBut again, lacking temperature specification, this question is
Rollout 3
/s. Alternatively, maybe exact problem has specific answer.\n\nAlternatively, maybe the problem uses R=8.
Rollout 7
answer for part(b) is N=24?\n\nWait, wait. Earlier we saw that in equations
Rollout 3
. Alternatively, given that room temp is common too.\n\nBut since there's no specified temperature, usually,
Rollout 3
, which would be 161 m/s.\n\nAlternatively, in some contexts, STP is
Rollout 3
67.2 m/s. Another possible result.\n\nSo after spending an hour on this, the possible
Rollout 3
likely leading to approximately 169 m/s.\n\nBut wait, perhaps my calculator computation precision was a
Rollout 3
is taken as 300 K for approximation.\n\nWait, given that no temperature is given, perhaps
Rollout 3
8K) so ~169 m/s.\n\nAlternatively, if the user is from somewhere particular:
Rollout 4
C_eq is 1206 pF.\n\nSimilarly, the minimum C_eq: C1=

UP_PROJ

Top 16 Positive Activations
Rollout 2
1000. As997 terms fromk=3 tok=999.
Rollout 0
? No, that's not correct. Each must have all primes exponents from 20! For
Rollout 2
09 and891 for each additional term fromk=3 onward.\n\nWait, but the real
Rollout 2
mod1000), then each subsequent term fromk>=3 is also-1 mod10
Rollout 7
"Find all possible values of N", but with given constraints.\n\nBut suppose that in this problem, the
Rollout 8
lengths of each block are variable (must be at least 1, obviously). But the key is that
Rollout 1
problem is over the Hilbert space, so thinking as in R^n, for which we know that even
Rollout 0
-free? No, that's not correct. Each must have all primes exponents from 20!
Rollout 1
only for the elements of S.\n\nTherefore, given that for all x ∈ S, ⟨x,
Rollout 1
, closed balls are weakly compact.\n\nBut given that spheres are closed and convex. Wait, the sphere
Rollout 1
can be formed that way. But the problem is given that such a set S exists, we need to
Rollout 7
getting promoted had original score64, which contributes to new score69, their total would be6
Rollout 1
ally embedded as such, and that the orthogonal complement required exists.\n\nBut since H is infinite-dimensional, this
Rollout 2
so the remainder is 0. However, given that our calculation says 109, which is
Rollout 0
does assigning a prime to a versus b affect the magnitude of a compared to b?\n\nWait, 2
Rollout 9
polygon. Wait, maybe not. Because when you say the overlapping region, the area within both triangle and
Top 16 Negative Activations
Rollout 1
vectors in $ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 7
participants is $N < 40$ . The passmark is fixed at $65
Rollout 1
helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$ be an infinite-dimensional
Rollout 7
of the repeaters $56$ . \nHowever, due to an error in the wording of
Rollout 2
$ is divided by $1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I
Rollout 1
$ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show
Rollout 1
points (not necessarily countable) in $ \mathcal{H}$ such that the distance
Rollout 6
that are of length $n$ . \n\n[list=1]\n [*] Given $k
Rollout 3
(Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the
Rollout 6
exists a polynomial $p : \mathbb{N} \mapsto \mathbb{N
Rollout 8
prime positive integers. Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's
Rollout 7
-promoted $47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to
Rollout 7
$ . The test results are\nthe following: \nThe average of all participants is $66
Rollout 9
to $EM$ in $\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 6
can decide the language $L$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay,
Rollout 1
a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$ be an infinite

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
Let me make sure.\n\nAccording to Maxwell-Boltzmann distribution, the mean speed is the average of
Rollout 3
(average) speed derived from the Maxwell-Boltzmann distribution:\n\n\[\nv = \sqrt{\
Rollout 3
8 K (unless stated otherwise), perhaps answer around \boxed{169} m/s. But
Rollout 2
of-1. However,-1^{997}= -1. So product mod125
Rollout 4
)/(a + b)^2 * Δa )2 + ( (a2)/(a + b)^
Rollout 1
. But with S possibly uncountable, this isn't feasible.\n\nAlternatively, take a countable subset
Rollout 1
answer\nGiven an infinite-dimensional Hilbert space \( \mathcal{H} \) and a set
Rollout 5
5. Compute the RMS: sqrt[(752 + 1752)/2] =
Rollout 2
mod1000. As997 terms fromk=3 tok=999.
Rollout 5
0\n\nThis is a quadratic in y: y2 -150y -1250
Rollout 1
’s commonly. But to have that the entire family's intersection is non-empty requires the family to have a
Rollout 2
.\n\nFirst, compute (-1)^{997} = -1 mod8, because 9
Rollout 1
helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$ be an infinite-dimensional
Rollout 7
the scores are increased, then the actual marks that count are the new ones. But the problem says pass
Rollout 5
sqrt[(752 + 1752)/2] = sqrt[(5625
Rollout 2
equivalent to multiplying by (-1)^{997}= -1. So total product ==9*
Top 16 Negative Activations
Rollout 1
two vectors would be sqrt( (d / sqrt(2))2 + (d / sqrt(2
Rollout 1
(x - y) ||·|| (sqrt(2}/d}{(x' - y}
Rollout 5
[150 ± 10*sqrt(725)] / 2 = 7
Rollout 5
2500)] / 2\n\nsqrt(72500) = sqrt(1
Rollout 1
distance each being d and radii d / sqrt(2}) would be non-empty? Not necessarily.
Rollout 1
sqrt(2))2 + (d / sqrt(2))2 ) = sqrt( d2/
Rollout 0
, a can vary from 1 up to sqrt(20!) if possible, but since 2
Rollout 5
/ 2 = [150 ± sqrt(72500)] / 2\n\n
Rollout 5
2 = 75 ± 5*sqrt(725)\n\nSince y = x +
Rollout 1
e_i} where each e_i is (sqrt(2)/d)(x_i - y) for
Rollout 0
2 < 20! or a < sqrt(20!). Therefore, the number of such a
Rollout 1
}, so r1 + r2 = d sqrt(2) �� 1.414
Rollout 1
our case, each sphere has radius d / sqrt(2} approx 0.707d
Rollout 1
infinite orthonormal system scaled by d / sqrt(2}, which is exactly the set S shifted by
Rollout 3
Rn at 298 K is sqrt(3RT/M). Let's compute:\n\n3*
Rollout 1
x' ∈ S, we have || (sqrt(2}/d}{(x - y) ||

Layer 14

GATE_PROJ

Top 16 Positive Activations
Rollout 7
50N/3 +59N/3=209N/3.\n\nSimilarly,
Rollout 7
/3, so 150N/3 +59N/3=209
Rollout 7
.\n\nNow, subtract 209N/3 from both sides:\n\n16x =71
Rollout 7
N/12 +162N/12 =224N/12=
Rollout 9
20y\n\nSo 20y = 100 => y = 5.\n\n
Rollout 4
10) / (2000 * (5000)) = (30,
Rollout 7
3 -14N=56N/3 -42N/3=14N
Rollout 7
=71N -209N/3= (213N -209
Rollout 7
9N/3=209N/3.\n\nSimilarly, the x terms:\n\n(75
Rollout 4
2 for V.\n\nSo, the relative error in E is sqrt( (0.0036
Rollout 4
as b for simplicity.\n\nSo, C_eq = (a*b)/(a + b). The partial derivative
Rollout 7
76*P + (A +5)*x) / (P +x) )=7
Rollout 7
56N/3\n\nSo,\n\nA*x -54x +18N =5
Rollout 7
N/12 =224N/12=56N/3, which matches
Rollout 7
50N to 150N/3, so 150N/3 +
Rollout 7
=71N\n\n205N/3 +32x=71N.\n\nSub
Top 16 Negative Activations
Rollout 0
/2 and N cannot come in pairs. So hence exponents for primes bigger than N/2 will
Rollout 2
which satisfies the first congruence.\n\nWait, hmm, that seems contradictory. Wait, no, according
Rollout 8
as arrangements where there are two contiguous color blocks. Hence, such sequences are called "linear arrangements with two
Rollout 7
that in equations, but led to impossibility. Thus, perhaps the answer is different.\n\nAlternatively, maybe
Rollout 0
as coprime factors, the way I'm thinking is that the assignment is of each prime as a
Rollout 0
must give it entirely to a or b, which leads to these coprime pairs. So seems that
Rollout 8
I think in probability problems with colored cards, unless specified, we need to consider each card as distinguishable
Rollout 1
as the graph of a linear operator or something. Hmm.\n\nAnother approach. Assume H has an orthon
Rollout 0
< b. But since N is fixed, that is a < N /a, so a^2
Rollout 1
sequence, but assuming that S is any set. Hmm.\n\nAlternatively, since you can extend any orthon
Rollout 9
10 max, which is inconsistent. So that suggests something's wrong.\n\nWait, perhaps my assumption is
Rollout 8
all green cards in the layout are adjacent (similarly if there are reds and greens, reds
Rollout 0
128?\n\nBut to confirm this, perhaps think again. The question is a bit more precise:
Rollout 6
oly, so the answer is invoking this theorem. Thus, the existence of such a sparse set S_L
Rollout 9
to EM, h can be anything. Let's think. If h is less than 10,
Rollout 1
same Hilbert space.\n\nWait, let me think again.\n\nGiven set S wherein between any two points,

UP_PROJ

Top 16 Positive Activations
Rollout 1
problem is over the Hilbert space, so thinking as in R^n, for which we know that even
Rollout 1
equations intersect at a common y.\n\nBut the problem now is to show that the system of equations ||x
Rollout 0
have a < b or a > b, so then to get the (a,b) with a <
Rollout 9
into two parts of 5 units each. But also, since EM is vertical, G is somewhere to
Rollout 1
≠ j. But actually no, in the previous, if we take multiple points, we need to ensure
Rollout 9
bottom-right and M is top-right, so that would be the vertical side. Wait, no, in
Rollout 1
x'||2 for any x, x', so then from the above equation, Re⟨x - x
Rollout 9
/h)(x -10), which rearranges to y = (-5/h)(x -10
Rollout 1
x, x', so then from the above equation, Re⟨x - x', y⟩ =
Rollout 5
) = 2/3. \n\nSo setting up the equation: (b + 25)/(
Rollout 8
G2 would not. But in the problem's terms, Kathy is only concerned with the colors. That
Rollout 3
1 m/s. However, if the problem is from a US textbook, some might take STP as
Rollout 1
⟩ = ||y||2. So substituting into ⟨x, x'⟩ - α - α
Rollout 7
5N = 71N. But also, the total score is equal to the sum of
Rollout 7
7(R -x)=71N.\n\nBut also, original equations hold:71P +5
Rollout 5
there were no algebraic mistakes. So the steps were:\n\n1. Found bases 75 and
Top 16 Negative Activations
Rollout 1
mutual distances are d, and the spheres have radius d / sqrt(2). Then, 2r
Rollout 6
are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $L \subseteq \{
Rollout 1
d. For example, if two points centers are d apart, spheres with radii r1 and r
Rollout 2
2=997 terms, each contributing a -1.\n\nSo first compute 9*99
Rollout 8
.\n\nWait, k=2: two sequences per k, contributing 2400. Wait,
Rollout 1
centers equals d2, radii squares are each d2 / 2. So sum of radii
Rollout 1
' contains 0 and other points each at distance d from 0 and distance d apart.\n\nBut this
Rollout 2
97 terms, each contributing a factor of -1 mod8.\n\nSo the product mod8 is (
Rollout 1
. Here, distance is d, and radius is d / sqrt(2). Since 2r =
Rollout 1
from each xi.\n\nEach sphere around xi with radius d / sqrt(2) must intersect.\n\nBut even
Rollout 8
1 to 4, then 2 sequences per k: RRGGG, GGRRR (for
Rollout 1
are at distance d from 0 and at distance d from each other.\n\nActually, no. When translating
Rollout 4
which as a percentage error would mean ±1.3% However, using error propagation method we found �
Rollout 1
property that all points except 0 are at distance d from 0 and from each other. So S
Rollout 1
which are therefore orthonormal when scaled up by sqrt(2)/d.\n\nTherefore, this direction shows
Rollout 1
Then the spheres around x and x' with radius d / sqrt(2) would intersect if and only

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
square speed, which is related to temperature and molar mass. Wait, but the question specifically says "
Rollout 3
the temperature in Kelvin, and M is the molar mass. Then there's the average speed, which
Rollout 3
Kelvin,\n- \( M \) is the molar mass of radon in kg/mol.\n\n**Steps
Rollout 2
9 -1). Therefore, the product is ��_{k=1}^{999}
Rollout 3
/mol. Let's check once again.\n\nThe molar mass of radon. Let me confirm.\n\nRad
Rollout 3
1 atm. At STP, the molar volume is 22.4 liters, which
Rollout 3
since it's a monatomic gas, the molar mass would be approximately 222 g/mol
Rollout 3
, Radon is a noble gas. Its molar mass is, since its atomic number is 8
Rollout 6
if the oracle confirms its presence.\n\n**Answer:** \nThe sparse set \( S \) consists of all
Rollout 3
For example, for nitrogen, which has a molar mass of 28 g/mol, the average
Rollout 3
they just want the formula expressed in terms of molar mass and R? No, the question is to
Rollout 3
/s.\n\nAs an assistant, I need to perhaps suggest both possibilities, but likely assume STP. Alternatively
Rollout 0
of unordered pairs is indeed 2^{k-1}, where k is primes. But, since prime
Rollout 0
this number is equal to 2^{k -1}, where k is the number of distinct prime factors
Rollout 4
3e-3, so 2.774 * 3.1623e
Rollout 3
Alternatively, maybe they expect us to recall the molar mass of radon and the temperature at STP
Top 16 Negative Activations
Rollout 3
the mean speed is called average speed and given by that formula with 8/π, while the rms
Rollout 1
d / sqrt(2) + d / sqrt(2) = d sqrt(2} ~1
Rollout 9
and GM.\n\nEquation for GE (from G(10 - h,5) to E(
Rollout 1
a sphere with center y and radius d / sqrt(2), then the necessary and sufficient condition is that
Rollout 5
oid into two smaller trapezoids? Each with height h/2, if the original height is
Rollout 1
a regular simplex in R^n with edge length sqrt(2), then the vectors from the centroid to the
Rollout 1
+ d / sqrt(2) = d sqrt(2} ~1.414 d >
Rollout 1
y } is orthogonal system with norm d / sqrt(2).\n\nFrom Rudin or another resource, there
Rollout 9
0? Wait, let's verify:\n\nFrom G(10 - h,5) to E(
Rollout 1
sqrt(2))2 + (d / sqrt(2))2 ) = sqrt( d2/
Rollout 1
centered at x ∈ S with radius d / sqrt(2)) is non-empty.\n\nBut how to check
Rollout 1
we need points at y's distance d / sqrt(2}, and vectors orthogonal.\n\nBut perhaps in this
Rollout 1
sum of radii l_s = d / sqrt(2) + d / sqrt(2) =
Rollout 1
to x and x' with distance d/sqrt(2), then the inner product is zero. So
Rollout 1
each lying on the sphere of radius d/sqrt(2) around y, and with pairwise inner product
Rollout 3
"mean molecular speed" is average speed, sqrt(8RT/(πM)).\n\nTherefore, assuming

Layer 15

GATE_PROJ

Top 16 Positive Activations
Rollout 1
are mutually orthogonal after scaling.\n\nWait, let me recall that if you have a regular simplex in R^n
Rollout 9
}\n\nWait, that seems correct.\n\nLet me check this.\n\nGiven that the area is 100
Rollout 5
the two bases. Wait, no, let me check. \n\nI remember that for areas divided by a
Rollout 1
two distinct vectors are orthogonal.\n\nFirst, let me recall that an orthonormal system requires two things:
Rollout 1
bert space, let $ d>0$ , and suppose that $ S$ is a
Rollout 5
5). Is that right?\n\nWait, let me check the calculations again step by step. Probably, I
Rollout 2
mod1000. But let me check this process again.\n\nFirst modulus8:\n\nProduct=(1
Rollout 4
0.88%. Wait, let me check the exact value.\n\nFirst, compute (ΔC
Rollout 1
the same Hilbert space.\n\nWait, let me think again.\n\nGiven set S wherein between any two points
Rollout 2
using the Chinese Remainder Theorem. Let me check if that's feasible.\n\nStarting with modulo 8
Rollout 5
/100. \n\nFirst, let me recall some trapezoid properties. In a tr
Rollout 1
to use the parallelogram law. Let me recall that in a Hilbert space, given two vectors
Rollout 8
<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to find the probability that when Kathy
Rollout 0
17, 19. Let me check. The primes less than or equal to 2
Rollout 3
so I need to be careful here. Let me check.\n\nI recall the root mean square (rms)
Rollout 0
.\n\nBut wait, can it be? Let me check with a smaller factorial.\n\nSuppose instead of
Top 16 Negative Activations
Rollout 2
;\n\nk=2,10^2 -1=99\n\nk=3, so1
Rollout 2
00\n\nSimilarly,10^k -1= -1 mod1000 fork>=
Rollout 2
k=3, so10^3 -1=999≡-1 mod10
Rollout 2
For k=1,10^1 -1=9;\n\nk=2,10^
Rollout 2
modulo 8 is \(2^k - 1\).\n - For \(k \geq
Rollout 2
9 is odd. 10^k-1 in general is written as (10-1
Rollout 2
8\), so \(2^k - 1 \equiv -1 \mod 8\).\n
Rollout 2
with k>=3 are10^k -1. For10^3=100
Rollout 2
00. Therefore,10^k -1= -1 mod1000 forall k
Rollout 0
. If the original rational number is in lowest terms, that means that a and b are copr
Rollout 2
-1). But (10^4 -1)=9999. 999
Rollout 2
Wait,\n\nWait, 10^k-1 for anyk is always even? No. When
Rollout 2
000, so10^4 -1=9999≡-1 mod1
Rollout 2
\n\nk=2:10^2 -1=99 mod125\n\nk=
Rollout 3
that's the rms speed. If they wanted the rms speed, but the question says "mean".\n\nAlternatively
Rollout 1
1, x2, x3 form an equilateral triangle with side length d. Suppose their positions are

UP_PROJ

Top 16 Positive Activations
Rollout 9
to $EM$ in $\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 5
exceed $x^2/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's
Rollout 8
prime positive integers. Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's
Rollout 1
a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$ be an infinite
Rollout 6
.\n<|im_start|>user\nLet's say a language $L \subseteq \{0,1\}
Rollout 2
$ is divided by $1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I
Rollout 6
that are of length $n$ . \n\n[list=1]\n [*] Given $k
Rollout 0
product of different primes, and how that relates? For example, consider assigning the small primes to a
Rollout 1
\nLet $ \mathcal{H}$ be an infinite-dimensional Hilbert space, let
Rollout 7
of promoted becomes79. To achieve this, need their total score:\n\ntotal promoted score =79
Rollout 6
as required.\n\nObviously, in an exam setting, need to explain more.\n\nAnother angle: Let S_L
Rollout 6
advice itself. Therefore, P_bad-angel would equal P.\n\nWait, let's formalize that.
Rollout 9
overlap, which is 80.\n\nTherefore, need to find h such that the area of overlap is
Rollout 4
C_eq max. To get maximum C_eq, need to max (C1*C2)/(C1
Rollout 1
helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$ be an infinite-dimensional
Rollout 9
, the answer is \boxed{25} . Seems done.\n\nThat's logical. Another way:\n\n
Top 16 Negative Activations
Rollout 1
translated condition.\n\nBut using the inner product formula:\n\nd2 = ⟨x - x', x - x
Rollout 1
In our case, r1 = r2 = d / sqrt(2}, so r1 + r
Rollout 7
24.\n\nOriginal promoted P=16, R=8.\n\nAfter increase promoted total:79
Rollout 2
=1:10 -1=9.\n\nk=2:100 -1=9
Rollout 1
must be an orthogonal set, with each vector having norm d / sqrt(2). Then scaling by sqrt
Rollout 1
the parallelogram law says that ||u + v||2 + ||u - v||2 =
Rollout 7
new averages are:\n\nNew average of promoted from original promoted:71 +5 =76. Wait
Rollout 1
(with center distance each being d and radii d / sqrt(2}) would be non-empty?
Rollout 7
N=36:\n\nP=24, R=12.\n\nx=36/1
Rollout 7
N.\n\nSame P=2N/3, R=N/3. So substituting:\n\n79
Rollout 6
each n, the number of strings in S of length log k + n is the sum over i=
Rollout 6
each n, the number of strings in S of length m = log k + n would be the sum
Rollout 9
square AIME, probably vertices in order A-I-M-E to make AIME. So when I originally
Rollout 5
apezoid with bases a and x and height k. The ratio of heights is k/h. The
Rollout 1
But the question is probably supposed to allow the System S to be mapped to orthonormal basis vectors.
Rollout 1
can embed S as y plus orthogonal vectors of norm d / sqrt(2}, then y exists. But

DOWN_PROJ

Top 16 Positive Activations
Rollout 4
696. The square root of that is sqrt(0.0000769
Rollout 1
cal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there exists
Rollout 4
7696\n\nTake the square root: sqrt(0.0000769
Rollout 0
20 prime factorization: 2^4 *3^2 *5^1. So number
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this math
Rollout 0
a and b, is 2^{k-1}. So for N =6, with three primes
Rollout 7
, does the passmark adjust with the scores? Hmm. In part (a) and (b)
Rollout 1
:\n\n• For each x ∈ S, ||√2/d (x - y) || = 1
Rollout 1
thonormal system. But howdoes that help.\n\nHmm.\n\nAlternatively, what if I use the following theorem
Rollout 3
3 K). Hmm.\n\nWait, to wrap my head around, let me check what standard temperature is commonly
Rollout 7
does the passmark adjust with the scores? Hmm. In part (a) and (b) the
Rollout 3
408.12\n\nSquare root: sqrt(28,408.12
Rollout 6
. [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these problems
Rollout 3
_rms = sqrt(3RT/M) = sqrt(3*8.314*2
Rollout 1
as d / sqrt(2) * sqrt(2) = d.\n\nTherefore, yeah, the setup
Rollout 3
298 K.\n\nV_rms = sqrt(3RT/M) = sqrt(3*
Top 16 Negative Activations
Rollout 6
and computed deterministically, it's like a P machine.\n\nSo unless NP = P, which is unknown
Rollout 6
oly theorem. The connection between sparse oracles and P/poly is established that a language is in P
Rollout 6
in complexity theory that P/poly is exactly the class of languages decidable by a polynomial-time machine with
Rollout 6
A(n).\n\nSo P_bad-angel is the class of languages decidable by a poly-time machine with
Rollout 6
's theorem equivalence between \( \textbf{P}/\text{poly} \) and sparse or
Rollout 3
, standard temperature for gas calculations is often 0°C, which is 273.15
Rollout 6
machine with a sparse oracle. (This is the Meyer, I think, theorem.) So perhaps this is
Rollout 6
, recall that P_angel is equivalent to P/poly, but with the advice strings being generated non
Rollout 6
^*$ is in $\textbf{P}_{angel}$ if there exists a polynomial
Rollout 6
oracle S_L. It is a standard result in complexity theory that P/poly is exactly the class of
Rollout 4
\n\nFirst, for capacitors in series, the equivalent capacitance C_eq is given by 1/C
Rollout 2
mod1000. But wait, hold on. Wait, 109 mod8 is
Rollout 0
7 has exponent 2? Wait for 7: floor(20/7)=2 (
Rollout 9
the top side of the square. Wait, hold on. \n\nWait, in the square AIME,
Rollout 0
a perfect square because, for example, exponent of 3 is 8 (even), exponent of
Rollout 6
_n for each input length n, similar to P/poly, but here the advice is fixed for all

Layer 16

GATE_PROJ

Top 16 Positive Activations
Rollout 0
?\n\nBut wait, an alternative approach: suppose we recognize that each coprime pair (a,b)
Rollout 2
891. Original problem:999 terms. First two terms,k=1,2.
Rollout 0
that is, a < sqrt(N!).\n\nBut for all possible assignments, the total number of cases where a
Rollout 1
\( d \) apart.\n - Translating the problem by a point \( y \) such
Rollout 7
*(R -x)=71N.\n\nSame P=2N/3, R=N/3
Rollout 7
all possible values of N. Let's check for each potential N (12,24,3
Rollout 2
invertible mod125. However, perhaps for k ≥3, 10^k is
Rollout 9
makes sense for an isosceles triangle with vertical base EM. Therefore, G is at (1
Rollout 0
=20!. Hence, equivalent to coprime pairs (a,b) with a*b=2
Rollout 1
.\n\nIf it's possible to find y such that translating by y makes the Gram matrix orthogonal, that would
Rollout 4
are independent and random, and combines them in quadrature, while the max-min approach assumes that all errors
Rollout 1
{ x - y : x ∈ S'} are orthogonal.\n\nWait, original problem: when translating back,
Rollout 8
, then the red is a single block (length 1) and the green is a single block (
Rollout 8
specified, we need to consider each card as distinguishable. Otherwise, the total number of possible sequences wouldn
Rollout 3
, that could vary.\n\nAlternatively, considering sometimes RMS speeds are reported. But no, the user specified "
Rollout 7
,N=36. Then, x=3 students who are original repeaters now promoted. If they
Top 16 Negative Activations
Rollout 1
2 - 3d2/4 - 3d2/4 + d2 / 2
Rollout 1
x'⟩ = 2d2 - 2⟨x, x'⟩.\n\nTherefore:\n\n⟨
Rollout 1
2 = 2||u||2 + 2||v||2.\n\nGiven that S is a
Rollout 1
0 ⇒ (2c - d2 - 2c + d2)/2 = 0 ⇒
Rollout 1
equations,\n\n2⟨x, y⟩ - 2⟨x', y⟩ = 2⟨x
Rollout 4
4)^2 )\n\nCalculating 2*0.004 = 0.008
Rollout 1
(2c - d2)/2 - (2α - ||y||2) = 0
Rollout 2
3, and 10 is 2*5, so 10 and 125
Rollout 4
2 + (2*0.004 )^2 )\n\nCompute:\n\n0.003
Rollout 1
2) - 3d2/4 -3d2/4 = d2 - (6
Rollout 4
06)^2 + (2 * 0.004)^2 )\n\nCalculating 2
Rollout 4
sqrt{3.6^2 + 2.4^2} = \sqrt{12
Rollout 4
5)^2 + (2 \cdot 0.004)^2} = \sqrt{
Rollout 2
helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when $9 \times 99 \times
Rollout 2
} \equiv 1 \times 3 \times (-1) \equiv -3 \equiv
Rollout 4
so 2.774 * 3.1623e-3 ��

UP_PROJ

Top 16 Positive Activations
Rollout 3
that standard problems involving heavy gases like radon typically compute the speed under standard conditions, either 0°C
Rollout 3
gas experiments, diffusion rates, Graham's law, typically at STP. But the molar mass is
Rollout 3
170? Due to perhaps answer key conventions. For purposes of being precise. If allowed as
Rollout 0
a subset S of the prime factors of N, hence product(a)= product of primes in S (with
Rollout 3
whether to use degrees Celsius or Kelvin. Wait, no, temperature must be in Kelvin in the formulae
Rollout 6
can have each block indexed by n and position, hence for each n, the number of strings is O
Rollout 3
15 K. Let me check.\n\nWait, possibly, the problem expects STP conditions (Standard Temperature
Rollout 4
and find the difference. But given that the problem provides the uncertainties as ± values, I think the standard
Rollout 8
Wait, wait, no. Wait, the problem says "shuffles the 10 cards and lays
Rollout 8
) but considering order.\n\nWait, but when they compute the probability, is the sample space the number of
Rollout 6
= |x|, which is poly(n) time), then run M(x, α_n). Since
Rollout 4
perhaps we need to compute maximum and min energy and find the difference. But given that the problem provides the
Rollout 6
think I'm going in circles here. Given the time constraints, maybe for problem 3, since it
Rollout 6
you can't encode solutions for all possible formulas of size n because SAT is NP-complete. Thus, likely
Rollout 7
't state a maximum score, but typically tests have maximum score like100. But original scores are
Rollout 7
59.\n\nWait, but actually, the problem says: "due to an error in the wording of
Top 16 Negative Activations
Rollout 0
are related to the divisors of 20! where each pair consists of two coprime numbers
Rollout 5
25, greatest integer not exceeding 181.25 is 181. Therefore
Rollout 0
prime powers in the factorization of 20! must go entirely to one of the numbers \(a
Rollout 0
of ways to write 20! as a product of two coprime numbers, \(a\)
Rollout 0
!. So they form a coprime pair whose product is 20!. Such pairs are related to
Rollout 5
\]\nThe greatest integer not exceeding 181.25 is 181.\n\n\
Rollout 0
the others are even. Thus, 20! is not a square because it has primes raised to
Rollout 0
rime, each prime factor in 20! must go entirely to \(a\) or entirely to
Rollout 0
:\n - The prime factors of 20! are 2, 3, 5,
Rollout 0
are even and odd.\n\nTherefore, 20! is not a perfect square because, for example,
Rollout 0
square. Therefore, the number of divisors N! So, accordingly, the number of such co-pr
Rollout 0
coprime pair whose product is 20!. Such pairs are related to the divisors of
Rollout 0
ization are all even? Wait, 20! has exponents:\n\nLet me do prime factorization
Rollout 0
1. **Prime Factorization of 20!**:\n - The prime factors of 2
Rollout 0
to rethink that.\n\nWait, actually 20! has prime factors with exponents. For example,
Rollout 0
1 will $20_{}^{}!$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>

DOWN_PROJ

Top 16 Positive Activations
Rollout 1
the equations are consistent.\n\nHence, the problem reduces to finding a t in H such that for all
Rollout 2
? 10 and 125 share factors: 5. Hmm. 125
Rollout 1
for all x ∈ S.\n\nThus, the problem reduces to finding y such that the function f(x)
Rollout 1
helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$ be an infinite-dimensional
Rollout 1
d / sqrt(2).\n\nTherefore, the problem reduces to showing that this intersection is non-empty. So
Rollout 0
have a < b. Let's check. Possible assignments:\n\n1. Assign none to a: a =
Rollout 1
having infinitely many such vectors. Thus, the problem reduces to finding the existence of such a configuration. So
Rollout 0
a < b, then we want the number of assignments where the product a < b. But since N
Rollout 0
the resulting numerator and denominator. For how many rational numbers between 0 and 1 will $20
Rollout 1
t = 2y.\n\nTherefore, the problem reduces to finding t in H and k ∈ R such
Rollout 0
, a < sqrt(N!).\n\nBut for all possible assignments, the total number of cases where a < b
Rollout 0
it?\n\nWait, but perhaps the counting is as follows: Each prime in 20!'s factor
Rollout 1
a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$ be an infinite
Rollout 0
0/2=10, 10 terms, then 5, 2, 1
Rollout 1
whole intersection is non-empty.\n\nTherefore, the problem reduces to showing that any finite subset of S has such
Rollout 0
. When we say that a and b are coprime with a*b=20!, then a
Top 16 Negative Activations
Rollout 7
) + (N/12)(A -79)=0\n\nSimplify:\n\n-2N
Rollout 3
or 298 K? Wait, standard temperature for gas calculations is often 0°C, which
Rollout 7
.\n\nTherefore:\n\n(N/12)(A -54) = 2N/3\n\nDiv
Rollout 3
6.512*(300 -7) = 66.512*
Rollout 2
25 mod8: 125 /8= 15*8=120
Rollout 3
possibly, the problem expects STP conditions (Standard Temperature and Pressure), which is 0°C and
Rollout 7
N/3\n\nFactor x:\n\nx(A -54) +18N =56N
Rollout 8
40=744/3024=744÷24=31
Rollout 3
Thus, gas at STP. Then STP temperature is 273 K. So maybe they
Rollout 0
1, b=720 (1 <720: yes)\n\n2. Assign 2
Rollout 1
/ sqrt(2) = d sqrt(2} ~1.414 d > distances between
Rollout 4
^-5) = 10^-2.5 �� 3.1623e
Rollout 7
4, average A, and the remaining R -x students with original scores <60, average B
Rollout 8
r! ) / (5 - (5 -r))! ; wait, maybe it's better to
Rollout 3
no temperature is given, the question must assume standard temperature, right? Because radon is a heavy gas
Rollout 7
score >=60 (since 60 +5 =65) are promoted. Therefore, in

Layer 17

GATE_PROJ

Top 16 Positive Activations
Rollout 3
(3RT/M), where R is the gas constant, T is the temperature in Kelvin, and M
Rollout 3
where:\n- \( R \) is the gas constant (\(8.314 \, \
Rollout 3
222 kg/mol.\n\nR is the gas constant, which is 8.314 J
Rollout 3
, maybe they expect us to recall the molar mass of radon and the temperature at STP.\n\n
Rollout 3
speed formula is sqrt(3RT/M), where R is the gas constant, T is the temperature in
Rollout 4
mathrm{~V}$. What is the percentage error in the calculation of the energy stored in this combination
Rollout 3
, textbook's choice. Given a chemistry book, Oxtoby, example problem: At 25
Rollout 3
if for some reason someone mixed them up.\n\nRMS speed for Rn at 298 K
Rollout 3
It's tricky.\n\nWait, given that molar mass is 222 (3 sig figs
Rollout 3
,\n- \( M \) is the molar mass of radon in kg/mol.\n\n**Steps:
Rollout 4
2 *10, etc. Let's recalculate the derivatives values more precisely:\n\nStarting with ΔC
Rollout 3
temperature in Kelvin, and M is the molar mass. Then there's the average speed, which is
Rollout 3
1 atm. At STP, the molar volume is 22.4 liters, which is
Rollout 3
, typically at STP. But the molar mass is key. So radon is often used in
Rollout 3
. Let me see. Let's do a quick estimation. Let's take 298 K first
Rollout 3
precise calculation using more decimal places.\n\nLet me recalculate the exact value.\n\n8 * 8.3
Top 16 Negative Activations
Rollout 0
of 20! must go entirely to one of the numbers \(a\) or \(b\).
Rollout 9
: y=(-1/5)(10)+2= -2 +2=0, GM:
Rollout 0
prime factor must go either to a or to b, since a and b are coprime. Therefore
Rollout 0
a,b) with a*b = N is equivalent to choosing a subset S of the prime factors of N
Rollout 0
factor, we assign it to either \(a\) or \(b\), but not both.\n\nTherefore,
Rollout 0
each prime can be assigned to either \(a\) or \(b\). But since each assignment corresponds to
Rollout 0
0!.\n\nEach prime factor must go either to a or to b, since a and b are copr
Rollout 9
0, y = (5/h)(0) + (10 - 50/h) =
Rollout 0
assignment corresponds to a unique pair (a,b) or (b,a), but we have \(a \
Rollout 6
), there is an entry (n, i) concatenated with the bit b. The length of each string
Rollout 1
, y⟩ = α for all x in S, which implies that y is orthogonal to the set S
Rollout 2
10^k -1)= (0 -1)= -1 mod8. Hence from k=
Rollout 0
of 20!.\n\nEach prime factor must go either to a or to b, since a and b
Rollout 9
5=(1/5)x +10 -2= (1/5)x +8.\n\nThus
Rollout 0
each prime is either assigned to a or to b, 2 choices per prime), but the assignment where
Rollout 2
(10^4 mod125 -1). 10^3=0 mod1

UP_PROJ

Top 16 Positive Activations
Rollout 1
sqrt(2) d �� 1.414d. The mutual distance between centers is
Rollout 3
\n\nWait no, perhaps with precise calculation using more decimal places.\n\nLet me recalculate the exact value.\n\n
Rollout 7
might have been misconsidered.\n\nWait, pass mark is fixed at65. Originally, promoted are
Rollout 3
/s}\n\]\n\n4. **Rounding to Significant Figures:**\n Given the inputs (molar
Rollout 3
For purposes of being precise. If allowed as per rounding rules, 168.5 rounds to
Rollout 4
) * Δb )^2 ]\n\nPlugging numbers:\n\na =2000, b=
Rollout 3
Wait no, perhaps with precise calculation using more decimal places.\n\nLet me recalculate the exact value.\n\n8
Rollout 7
is still 65? Wait, but pass mark is fixed at 65. Wait, the
Rollout 3
}\n\]\n\n4. **Rounding to Significant Figures:**\n Given the inputs (molar mass
Rollout 1
simplex, in 3D, i.e., tetrahedron, but even a triangle is
Rollout 4
8 is okay. Alternatively, let's calculate exact decimals:\n\nComputed ΔC_eq / C_eq =
Rollout 1
of completing the system iterating over the points by transfinite induction. (Dragon a bit here.)\n\nAlternatively,
Rollout 8
which is 7440, and check over a smaller case. Suppose instead that Kathy had
Rollout 4
4% error. \n\nBut since combining them by quadrature gives sqrt(0.52 +0
Rollout 3
perhaps they just want a ballpark figure, maybe expecting answer ~170 m/s. Alternatively, maybe
Rollout 1
^n, for which we know that even for three points, you can find such a y. But in
Top 16 Negative Activations
Rollout 7
Similarly, the x terms:\n\n(75 -59)x=16x.\n\nTherefore, the
Rollout 7
3 -18N= (56 -54)N/3 ?\n\nWait, wait:\n\n
Rollout 0
), b=45 (3^2 *5). gcd(16,45)=1
Rollout 2
, each term (10^k -1) mod 8 is (2^k -1
Rollout 8
00) + (2400 +2400) = 2400
Rollout 2
8, hence (10^k -1)= (0 -1)= -1 mod8.
Rollout 9
no, if the altitude is from G to EM, which is length h, then if h is the
Rollout 8
7200. 240 +7200= 7440.
Rollout 7
Total promoted-from-repeaters=448 -252=196. So A=
Rollout 9
= 100 - 20y + y^2\n\nSubtract y^2 from
Rollout 7
5N/3=(213N -205N)/3=8N/3
Rollout 7
5P) + (56R +5R) = 76P + 6
Rollout 0
19). Hence, 2^{8 -1}= 128. Hence, result.
Rollout 0
6*45=24^2*5).\n\nBut in general, n! will have prime
Rollout 2
0 mod8, so (2^k -1) �� -1 mod8.\n\nTherefore,
Rollout 2
1). But (10^4 -1)=9999. 9999

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
(3RT/M), where R is the gas constant, T is the temperature in Kelvin, and M
Rollout 0
in N's prime factorization), which requires multiplicativity. For each prime p | N, assign all
Rollout 3
222 kg/mol.\n\nR is the gas constant, which is 8.314 J
Rollout 3
where:\n- \( R \) is the gas constant (\(8.314 \, \
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a
Rollout 8
specified, we need to consider each card as distinguishable. Otherwise, the total number of possible sequences wouldn
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides
Top 16 Negative Activations
Rollout 7
5P = 10R\n\nDivide both sides by 5:\n\nP = 2R
Rollout 5
)] = 125H\n\nDivide both sides by H (assuming H ≠ 0):\n\n
Rollout 5
150; subtract 2b from both sides: b + 75 = 1
Rollout 8
.\n\nSimplify this fraction.\n\nFirst, divide numerator and denominator by 240: numerator becomes
Rollout 2
1)=109,000 -109=108,891
Rollout 9
y^2\n\nSubtract y^2 from both sides:\n\n0 = 100 -
Rollout 7
66P and 56R from both sides:\n\n5P = 10R\n\n
Rollout 5
) = 125H\n\nDivide both sides by H (assuming H ≠ 0):\n\n
Rollout 4
))) ΔC1 ]^2 \n\nMultiply numerator and denominator:\n\n= [ (C2 ΔC1
Rollout 1
2 + ||x'|||^2).\n\nDividing both sides by 2:\n\n-⟨x + x
Rollout 5
25x + 9375 from both sides:\n\nx^2 - 181
Rollout 7
) = 2N/3\n\nDivide both sides by N (assuming N ≠0, which
Rollout 7
, subtract 209N/3 from both sides:\n\n16x =71N -
Rollout 5
This is a quadratic in y: y2 -150y -12500 =
Rollout 5
. In a trapezoid, the midline (the segment connecting the midpoints of the legs
Rollout 8
240= Let's check by dividing numerator and denominator by 240: 74

Layer 18

GATE_PROJ

Top 16 Positive Activations
Rollout 5
be b, then the longer base is b + 100. The midline length is (
Rollout 3
24.942*298 ��24.942*3
Rollout 5
b + 50) + (b + 100))/2 * (h/2
Rollout 3
66.512 * 298:\n\nLet me compute this multiplication.\n\n66.
Rollout 5
is (b + 25)/(b + 75) = 2/3. \n\n
Rollout 3
8*8.314*298= about 19,820 (
Rollout 3
66.512 * 298 = Let me compute 66.51
Rollout 2
since 1000 is 8 * 125 and 8 and 12
Rollout 3
8.314 * 298\n\nFirst, compute 8*8.3
Rollout 3
66.512*298 �� 19,820.
Rollout 4
C1 / C1 = 10 / 2000 = 0.00
Rollout 3
66.512 * 298 = let's compute 66.51
Rollout 3
14 (4 sf) * 298 (3 sf) / 0.22
Rollout 3
3*8.314*298=3*8.314*2
Rollout 3
8.314 * 298) / (π * 0.22
Rollout 3
8*8.314*298)/(3.1416*0.
Top 16 Negative Activations
Rollout 1
$ is a set of points (not necessarily countable) in $ \mathcal{H
Rollout 1
is a set of points (not necessarily countable) in $ \mathcal{H}$
Rollout 9
0 (base EM) and two lines intercepting x=0. Probably the overlap inside is the tr
Rollout 6
x of length n, needs to get the bits of α_n, which is p(n) long,
Rollout 6
length of each string is O(log n + log i). Since p(n) is polynomial in n,
Rollout 9
, GM intersects the left edge at (0,10 -50/h). Similarly, these are
Rollout 9
from (0,0) to (0,10)). Since G is to the left of the
Rollout 2
the number of such terms, if it's odd number, it's109, else89
Rollout 9
h,5) to E(10,0), slope is -5/h. So equation can
Rollout 1
in H. If S is a set of points each lying on the sphere of radius d/sqrt(
Rollout 1
, origin with one of them (z = 0). Wait, if x0 is in S,
Rollout 0
\nGiven a rational number, write it as a fraction in lowest terms and calculate the product of the resulting
Rollout 6
that seems impossible if we need to get the bits of α_n from the oracle. Because a query to
Rollout 1
2. Then the set is a collection of vectors each of length d, with pairwise inner product d2
Rollout 1
all points lie on a sphere centered at \( y \) with radius \( \frac{d}{\
Rollout 0
1, we need to write it as a fraction in lowest terms \(\frac{a}{b

UP_PROJ

Top 16 Positive Activations
Rollout 0
2.\n\nWait, 7^2 divides into 20! but 7^3 does
Rollout 8
4=31; 3024÷24=126. So yes,
Rollout 8
3, etc. 126 is 2* 63=2*7*
Rollout 8
=31, 30240÷240=126. As above
Rollout 2
125. 1000 divided by125 is8, so 10
Rollout 2
5. Hmm. 125 is 5^3, and 10 is
Rollout 3
square root of 28,408.12:\n\nsqrt(28,40
Rollout 4
8%. Since 0.877 is 0.88 when rounded to two decimal places
Rollout 8
1; 3024÷24=126. So yes, 31
Rollout 0
of the total pairs, so 256/2 =128. Hence 12
Rollout 8
240: 7440 divided by 240. Let's see:
Rollout 8
0240. 7440/30240= Let's check by
Rollout 4
compute that more accurately.\n\n4.327 / 1200. Let me calculate
Rollout 7
294. Which is 294/3=98 per student, but it's
Rollout 4
�0.88%, which is half of 1.76% (max-min). But
Rollout 0
is not square-free (as 720=16*45=24^2
Top 16 Negative Activations
Rollout 4
are independent and random, and combines them in quadrature, while the max-min approach assumes that all errors
Rollout 4
is about ±1.3%, whereas the quadrature gives ~0.88%, the answer would
Rollout 5
solve this problem about a trapezoid with bases differing by 100 units. The segment
Rollout 5
height k creates a smaller trapezoid with bases a and x and height k. The ratio of
Rollout 9
. But that seems like trapezoid with bases at x=0 (the two points: (
Rollout 4
then compute the maximum and minimum possible energy considering the uncertainties and find the percentage difference. But error propagation using
Rollout 9
lines is the area of trapezoid with bases at x=0 (length 8-2
Rollout 4
errors are independent and random, and combines them in quadrature, while the max-min approach assumes that all
Rollout 3
average) speed derived from the Maxwell-Boltzmann distribution:\n\n\[\nv = \sqrt{\frac
Rollout 3
. Let me make sure.\n\nAccording to Maxwell-Boltzmann distribution, the mean speed is the average
Rollout 4
the two 0.5% errors via quadrature with coefficients. Then the total energy error is sqrt
Rollout 3
me make sure.\n\nAccording to Maxwell-Boltzmann distribution, the mean speed is the average of the
Rollout 9
10 units. Let's label the square with points A, I, M, and E. Probably
Rollout 3
for the mean (average) speed derived from the Maxwell-Boltzmann distribution:\n\n\[\nv =
Rollout 5
Let me denote the original trapezoid with bases a = 75, b = 1
Rollout 1
helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$ be an infinite-dimensional

DOWN_PROJ

Top 16 Positive Activations
Rollout 2
)...(10^{999} -1). We need to compute P mod 10
Rollout 0
a = product_{p in S} p^{e_p}, and b = 20! /
Rollout 2
891*(1000 -1)=891,000 -8
Rollout 4
C1 / C1 = 10 / 2000 = 0.00
Rollout 2
is k=1 (10^1 -1), the second is k=2 (10
Rollout 2
=891*(1000 -1)=891*1000 -
Rollout 0
, with three primes, 2^(3 -1)=4, which matches what we saw for
Rollout 1
cal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there exists
Rollout 5
be b, then the longer base is b + 100. The midline length is (
Rollout 0
19). Hence, 2^{8 -1}= 128. Hence, result.
Rollout 8
=1: [5! / (5 -1)! ] = 5! / 4!
Rollout 2
99 (10^999 -1). Therefore, the product is ��_{k
Rollout 8
the number of possible sequences is 10 * 9 * 8 * 7 * 6
Rollout 2
Thus, each term (10^k -1) mod 8 is (2^k -
Rollout 4
985= (2000 -10)*(3000 -15)=
Rollout 2
)*(...)*(10^999 -1). So total number of terms:999
Top 16 Negative Activations
Rollout 8
7440/30240= Let's check by dividing numerator and denominator by
Rollout 8
7440/30240=744/3024=
Rollout 5
.5H.\n\nSo we have two equations:\n\n1. (75 + x)h =
Rollout 8
the number of ordered arrangements. For example, for R^r G^{5−r}: we need
Rollout 9
/h <5, so (0,50/h) is below (0,5).\n\n10
Rollout 8
6.\n\nTherefore, 31/126. Then check if they can be simplified. Since
Rollout 8
As above. Therefore 31/126.\n\nBut I wonder, was that correct? Let
Rollout 5
algebraic mistake.\n\nStarting from the two equations:\n\n1. (75 + x)h =
Rollout 8
, the probability is 31/126, so m=31, n=1
Rollout 8
26. So 31/126.\n\nWait, wait: 7440
Rollout 7
N)/12= 3N/12= N/4 students. Their average is
Rollout 8
only be two blocks, so the layout is either R...RG...G or G...GR...R
Rollout 8
greens are together. So such sequences would look like RRRGG, RGGGG, GGRRR
Rollout 9
/h)(x -10) = (-5/h)x + (50/h). So yes.\n\n
Rollout 8
. So yes, 31/126≈0.246.\n\nTo verify if
Rollout 8
240. However, in the color sequence RRRRR, how many ordered hands correspond to that

Layer 19

GATE_PROJ

Top 16 Positive Activations
Rollout 1
a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$ be an infinite
Rollout 9
helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides of length $10$ units
Rollout 9
mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides of length $10$ units.
Rollout 2
times 999 \times \cdots \times \underbrace{99\cdots
Rollout 1
mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$ be an infinite-dimensional Hil
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides of
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red cards
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants is
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $L
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values $
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in the
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number, write it as
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when $9 \times
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a trapezoid
Rollout 9
0$ units. Isosceles triangle $GEM$ has base $EM$ , and
Top 16 Negative Activations
Rollout 4
�0.88%, which is half of 1.76% (max-min). But
Rollout 1
sphere has radius d / sqrt(2} approx 0.707d. So sum of
Rollout 1
that for any x, ||v_x|| = 1, and for any x ≠ x', ⟨
Rollout 1
⟨v_x, v_{x'}⟩ = 0.\n\nCalculating the norm of v_x:
Rollout 8
the number of possible sequences is 10 * 9 * 8 * 7 * 6
Rollout 7
is 75. Wait, that’s not 76. So even though original promoted had their
Rollout 2
k -1 is odd. Because even minus1 is odd. Therefore, every term in the product is
Rollout 4
606% as relative error.\n\nWait, 0.36058333
Rollout 8
each green card identical? Because if that's the case, then a sequence is determined by the colors.
Rollout 2
-1 is odd. Because even minus1 is odd. Therefore, every term in the product is odd
Rollout 0
), so that's exponent 2.\n\nWait, 7^2 divides into 20! but
Rollout 0
of unordered pairs would be \(2^{8 - 1} =128\), since each
Rollout 2
nines. So the product is 9 * 99 * 999 * ... *
Rollout 1
than x - y and x' - y are orthogonal, we can create infinite such orthogonality
Rollout 0
(all primes assigned to b) and b = 20! (there's exactly one such case
Rollout 2
4 = 0 mod 8\n\nWait, 2^3 = 8 �� 0

UP_PROJ

Top 16 Positive Activations
Rollout 9
, y). But the triangle being isosceles with base EM. Since triangle GEM is is
Rollout 5
(18125) = sqrt(25 * 725) = 5
Rollout 9
5)\) because the triangle is isosceles with \(GE = GM\).\n\n4. **
Rollout 9
y. But as the triangle is isosceles with GE = GM.\n\nCalculating GE distance:\n\n
Rollout 9
EM. Since triangle GEM is isosceles, meaning legs GE and GM are equal. Wait
Rollout 9
equal. Wait, if it's isosceles with base EM, then sides GE and GM must
Rollout 5
18125, so x = sqrt(18125) = sqrt(2
Rollout 9
and M. Since the triangle is isosceles with base EM, the apex is G. And
Rollout 9
forms the triangle. The triangle is isosceles with base EM, so the legs would be GE
Rollout 4
F\n\nThus, ΔC_eq =sqrt(3.62 +2.42)=sqrt
Rollout 3
408.123\n\nsqrt(28,408.123)=
Rollout 1
of radii l_s = d / sqrt(2) + d / sqrt(2) = d
Rollout 1
||x - x'|| = d = sqrt(2) * ||x - y|| / sqrt{
Rollout 3
rounded up. However, in calculator, sqrt(28408.12)= 1
Rollout 0
< 20! or a < sqrt(20!). Therefore, the number of such a is
Rollout 9
=10), and is an isosceles triangle, then point G (the apex) is
Top 16 Negative Activations
Rollout 2
remainder when the product of numbers like 9, 99, 999, and so
Rollout 5
segment connecting the midpoints of the legs divides the trapezoid into two regions with areas in the
Rollout 0
is some fraction as part of between 0 and 1. When written in lowest terms, numerator a
Rollout 5
which is the length of the segment that divides the trapezoid into two equal areas, and then
Rollout 8
5, those could be anywhere from 0 to 5 reds. Wait, but in the problem
Rollout 8
GG, etc. But something like RRRGR wouldn't work because the reds are not all together
Rollout 0
Let's ask for rationals between 0 and 1 with product a*b=720,
Rollout 2
a number with 999 nines, is divided by 1000. Hmm,
Rollout 2
etc. Wait,\n\nWait, 10^k-1 for anyk is always even? No
Rollout 2
perhaps for k ≥3, 10^k is 0 mod125?\n\nWait,
Rollout 2
anyk>=1. Then 10^k -1 is odd. Because even minus1 is
Rollout 2
2 mod 8, so 10^k mod 8 is 2^k mod
Rollout 0
a < b or a > b, so then to get the (a,b) with a < b
Rollout 5
is a segment parallel to the bases that divides the trapezoid into two regions of equal area,
Rollout 0
.\n\nFirst, a rational number between 0 and 1 can be written as a fraction \(\frac
Rollout 2
\nTo find the remainder when \(9 \times 99 \times 999 \times

DOWN_PROJ

Top 16 Positive Activations
Rollout 1
||2 = 2||u||2 + 2||v||2.\n\nGiven that S is
Rollout 8
the number of possible sequences is 10 * 9 * 8 * 7 * 6
Rollout 1
1) each vector has norm 1, and 2) any two distinct vectors are orthogonal.\n\nGiven
Rollout 5
be b, then the longer base is b + 100. The midline length is (
Rollout 5
is a segment parallel to the bases that divides the trapezoid into two regions of equal area,
Rollout 0
.\n\nFirst, a rational number between 0 and 1 can be written as a fraction \(\frac
Rollout 1
that for any x, ||v_x|| = 1, and for any x ≠ x', ⟨
Rollout 4
1/C_eq = 1/C1 + 1/C2. So, I can write that
Rollout 2
remainder when the product of numbers like 9, 99, 999, and so
Rollout 0
.\n\nBut since the fraction is between 0 and 1, \(a < b\). So for
Rollout 5
which is the length of the segment that divides the trapezoid into two equal areas, and then
Rollout 9
10, h is from 0 to 10? But EM is the vertical edge of
Rollout 2
nines. So the product is 9 * 99 * 999 * ... *
Rollout 1
⟨v_x, v_{x'}⟩ = 0.\n\nCalculating the norm of v_x:
Rollout 6
not the other way.\n\nWait, but the problem doesn't say that k is given as part of the
Rollout 0
Let's ask for rationals between 0 and 1 with product a*b=720,
Top 16 Negative Activations
Rollout 3
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 0
$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how
Rollout 2
. Hence, answer109.\n\nTherefore, boxed answer is \boxed{109}\n\n**
Rollout 0
not a square. Therefore, the number of divisors N! So, accordingly, the number of such
Rollout 8
$m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to
Rollout 4
this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage
Rollout 2
1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here:
Rollout 5
/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to
Rollout 9
$\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry
Rollout 1
, x2 ∈ S (with ||x1 - x2|| = d), the linear equations ⟨
Rollout 0
with a unique pair (a,b) where a < b. So the count is half the number of
Rollout 3
169 m/s. Thus, present answer boxed as \boxed{169 \text{
Rollout 4
... more like 0.88%. So box this as 0.88%.\n\nAlternatively,
Rollout 3
169 m/s.\n\nTherefore the valid boxed answer is \boxed{169}.\n\n
Rollout 1
- ||x2||2 ⇒ ⟨x1 - x2, y⟩ = (||x1
Rollout 0
of such a is half the total number of divisors, adjusted for square factors. However, we are

Layer 20

GATE_PROJ

Top 16 Positive Activations
Rollout 8
are the only possibilities for mixed, and each is already a block of 1 red and 1 green
Rollout 0
translates as, (since 20! is fixed and a*b=20!), how many ways
Rollout 0
b?\n\nWait, 20! is a fixed number, but how is a related to b?
Rollout 7
must be a common multiple of 3 and 12, that is, a multiple of 1
Rollout 1
between centers equals d2, radii squares are each d2 / 2. So sum of radi
Rollout 8
the only possibilities for mixed, and each is already a block of 1 red and 1 green.
Rollout 1
are governed by\n\nFor any two spheres, there is a non-empty intersection because the distance squared between centers
Rollout 0
?\n\nWait, 20! is a fixed number, but how is a related to b? a
Rollout 0
in lowest terms, numerator a and denominator b must satisfy that. However, a different way to phrase this
Rollout 0
as, (since 20! is fixed and a*b=20!), how many ways can
Rollout 7
is 294/3=98 per student, but it's impossible since max original repe
Rollout 6
i ranges up to p(n), which is polynomial, log i is O(log n). Hence, each
Rollout 1
each sphere has radius d / sqrt(2} approx 0.707d. So sum
Rollout 7
to 64.\n\nBut A has to be such that A is an integer (since all individual scores
Rollout 8
the blocks are adjacent, the sequence color is fixed once you choose which color comes first. Therefore, for
Rollout 0
a,b) with a < b corresponds to the exact opposite pair (b,a) with b > a
Top 16 Negative Activations
Rollout 3
. Let's check with calculator steps.\n\nAlternatively, sqrt(28,408). Let's
Rollout 5
, and then find the greatest integer not exceeding x2/100. \n\nFirst, let me
Rollout 3
7≈0.047. Thus, sqrt≈168.5 +0.0
Rollout 4
approx 2.7749, so sqrt(7.7e-5)=2.
Rollout 3
28,408.12:\n\nsqrt(28,408.12
Rollout 4
7 is 7.7e-5. sqrt(7.7e-5). Well,
Rollout 3
= 28,561. So sqrt(28,436) is approximately
Rollout 5
= 18125. Then x2 /100 = 1812
Rollout 3
=15.87.\n\nLinear approximation:\n\nsqrt(x + dx) �� sqrt(x) +
Rollout 4
are maximum.\n\nWait, but because (C1*C2)/(C1 + C2), if both
Rollout 3
8,408.12. So sqrt(28,408.12
Rollout 2
125. Therefore, (10^k -1) �� -1 mod12
Rollout 4
(7.7e-5). Well, sqrt(7.7) is approx 2.
Rollout 1
for each x in S, v_x = (sqrt(2)/d)(x - y). Then
Rollout 3
2·mol·K). Then, 8*R*T has units (kg·m2)/(s
Rollout 3
sqrt(x)=168.5.\n\nSo sqrt(28,392.25

UP_PROJ

Top 16 Positive Activations
Rollout 3
69 m/s.\n\n**Final Answer**\n\boxed{170} m/s\n\nWait no
Rollout 3
3, so if need to answer, in the absence of the temperature in the problem, I should either
Rollout 3
.5 m/s. If using calculator accurate value.\n\nSo the answer is approximately 168.
Rollout 3
69} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon gas is \boxed
Rollout 1
problem statement.\n\n**[Final Answer]**\n\boxed{\text{Such a point } y \text
Rollout 4
0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of the energy stored is
Rollout 3
) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular speed
Rollout 9
25.\n\nPerfect.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(A
Rollout 3
169 m/s. Thus, present answer boxed as \boxed{169 \text{
Rollout 3
So all steps are correct.\n\nThus, conclusion: at 298 K, the mean molecular speed
Rollout 3
**\nThe mean molecular speed of radon gas is \boxed{170} m/s.\n\nWait
Rollout 4
\boxed{0.88\%}.\n\n<|im_start|>answer\nTo find the percentage error in the
Rollout 0
}=128.\n\n**Final Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven a
Rollout 3
answer approximately 170 m/s. But given compute precisely, it is about 168
Rollout 3
**\nThe mean molecular speed of radon gas is \boxed{169} m/s.\n\n<|im_start|>
Rollout 3
168.5 m/s.\n\nIn absence of more specifications, but given standard calculations in textbooks
Top 16 Negative Activations
Rollout 2
\geq 3\), \(2^k \equiv 0 \mod 8\),
Rollout 2
=3. For k>=3:2^k mod8. But 2^3=8
Rollout 2
Thus, for k ≥3, 2^k �� 0 mod8, so (2
Rollout 2
perhaps for k ≥3, 10^k is 0 mod125?\n\nWait,
Rollout 2
k=10^{3}*10^{k-3}≡0*10^{k
Rollout 2
0^k mod 8 is 2^k mod 8.\n\nThus, each term (1
Rollout 2
is0 mod8, hence (10^k -1)= (0 -1)= -1
Rollout 2
0^k -1)≡(2^k -1). For k=1:2-
Rollout 2
maybe see if we can find when (2^k -1) �� 0 mod 8
Rollout 2
for k>=3, hence (10^k -1)≡-1 mod125
Rollout 2
k≥3: terms are 10^k -1 �� -1 mod125
Rollout 2
9s can be written as 10^n - 1. For example:\n\n- 9
Rollout 2
� 0 mod8, so (2^k -1) �� -1 mod8.\n\n
Rollout 7
P +56(N -P)=66N\n\nExpand:71P +56N
Rollout 2
≡2 mod8, so (10^k -1)≡(2^k -1
Rollout 7
→-3P +A x -79x=0\n\n→-3P +x(A

DOWN_PROJ

Top 16 Positive Activations
Rollout 1
<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there exists a point y in the
Rollout 8
the number of possible sequences is 10 * 9 * 8 * 7 * 6
Rollout 1
.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there exists a point y in
Rollout 9
<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry problem where there's a square
Rollout 9
$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry problem where there's
Rollout 1
}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there exists a point y
Rollout 8
m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to find
Rollout 5
100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve
Rollout 0
.\n\nBut since the fraction is between 0 and 1, \(a < b\). So for
Rollout 9
triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry problem
Rollout 2
nines. So the product is 9 * 99 * 999 * ... *
Rollout 8
m$ and $n$ are relatively prime positive integers. Find $m + n$ .\n<|im_start|>assistant
Rollout 0
256\), but each pair is counted twice except when \(a = b\). But
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this math
Rollout 0
be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how many
Rollout 5
be b, then the longer base is b + 100. The midline length is (
Top 16 Negative Activations
Rollout 3
of some standard conditions. Hmm.\n\nWait, in kinetic theory, mean speed depends on temperature, so we
Rollout 4
percentage error in the calculation of the energy stored is \boxed{0.88\%}.\n\n
Rollout 3
298 (3), the least number of sig figs is 3, so result should be
Rollout 9
, the problem says the base of the triangle is EM, which in this case is the vertical side.
Rollout 3
hasn't specified the temperature, this seems ambiguous. However, in many standard problems, they might take
Rollout 3
. However, in some contexts, "mean molecular speed" might refer to rms. Wait, I need
Rollout 1
d / sqrt(2), and the vectors from y to each x must be pairwise orthogonal.\n\nSo the
Rollout 3
people use mean and average interchangeably, but in kinetic theory, they are different. Let me make sure
Rollout 3
, the mean molecular speed of radon gas is \boxed{170} m/s if we
Rollout 1
norms of the points in S are the same. However, in a Hilbert space, we can shift
Rollout 2
can find the remainder. However, calculating this directly seems impossible because the numbers are huge. So, perhaps
Rollout 3
no, the problem says "compute the mean molecular speed," so they must want a numerical answer. Therefore
Rollout 1
2).\n\nTherefore, the problem reduces to showing that this intersection is non-empty. So, since each sphere
Rollout 2
-3≡5 mod8.\n\nOkay, modulus8 seems fine. Good.\n\nNow modulus125
Rollout 3
there's the average speed, which is sqrt(8RT/(πM)). The problem mentions "mean
Rollout 3
for the mean (average) speed derived from the Maxwell-Boltzmann distribution:\n\n\[\nv =

Layer 21

GATE_PROJ

Top 16 Positive Activations
Rollout 0
then the number of coprime pairs with a < b is exactly half of the total number of cop
Rollout 0
factors, number of coprime pairs with a < b and a*b=20! is
Rollout 6
S_i has at most p(n) strings of length n, total is k*p(n). Since k
Rollout 6
_i has at most p(n) strings of length n, total is k*p(n). Since k is
Rollout 1
of closed sets with the finite intersection property is non-empty if the space is compact. However, Hilbert
Rollout 0
factors. The number of coprime pairs a < b where a*b=720 is
Rollout 3
, I need to calculate the mean molecular speed of radon gas. Hmm, first, I remember there
Rollout 1
verify, this requires that every finite intersection is non-empty.\n\nLet's take as finite subset, say F
Rollout 2
mod8\n\nSubtract 5:\n\n5m≡0 mod8\n\nThus, 5m≡
Rollout 6
each n, the number of strings in S of length log k + n is the sum over i=
Rollout 3
divided by (kg/mol) is (m2/s2). Square root is m/s. So unit
Rollout 6
), that corresponds to one string of length m per n equal to p^{-1}(m). However,
Rollout 2
�5 mod8, hence 5m ��0 mod8, leading to m≡0 mod
Rollout 0
primes), number of coprime pairs with a < b=4=2^{3-1}
Rollout 1
metric space. However, the intersection might still be empty unless certain conditions hold, such as the family of
Rollout 0
coprime pair (a,b) with a < b corresponds to the exact opposite pair (b,a
Top 16 Negative Activations
Rollout 5
but these are not similar necessarily. \n\nAlternatively, think algebraically. Let me denote the original trape
Rollout 8
are adjacent and all greens are adjacent.\n\nAlternatively, think of the problem as the laid-out 5 cards
Rollout 2
of1000 here?\n\nAlternatively, maybe because many of the terms in the product from k=
Rollout 1
system.\n\nBut how does this help?\n\nAlternatively, think of y as a solution to the system of equations
Rollout 3
\boxed{169}.\n\nBut initially I thought 168.5 rounded to
Rollout 4
figures, 0.88%.\n\nAlternatively, maybe give it as 0.877%,
Rollout 5
step. Let me verify quickly. \n\nAlternatively, perhaps it is quicker to recall the formula for the line
Rollout 2
multiple of1000 here?\n\nAlternatively, maybe because many of the terms in the product from k
Rollout 9
. Since each side is 10 units. So AI would be vertical, IM horizontal, ME vertical
Rollout 8
color sequences.\n\nTherefore, if color sequences are counted based on color orders, there are 10 happy
Rollout 7
. But how does this make sense?\n\nAlternatively, perhaps during the problem, the composition of promoted and repe
Rollout 3
boxed{169}.\n\nBut initially I thought 168.5 rounded to 1
Rollout 9
etc. Wait, perhaps getting complicated.\n\nAlternatively, maybe split the overlapping area into the part from x=
Rollout 1
requires transfinite induction.\n\nAlternatively, if we realize that the set S, with all mutual distances d,
Rollout 4
when rounded to two decimal places. \n\nAlternatively, maybe exact calculation. Let's do exact calculation.\n\nWe
Rollout 7
and in part(b) equations require impossible averages, leading no N satisfies the given.\n\nThus:\n\n(a)

UP_PROJ

Top 16 Positive Activations
Rollout 6
the machine M, given x of length n, needs to get the bits of α_n, which is
Rollout 7
total promoted score =79*(P +x).\n\nBut original promoted now scores higher. New scores=
Rollout 7
+x)=79*(16 +2)=79*18=1422
Rollout 7
-promoted total=47*(R -x). Then:\n\n79(P +x) +4
Rollout 4
2*0.004 )^2 )\n\nCompute:\n\n0.003605
Rollout 1
+ ||x'||2 - d2)/2 - ⟨x + x', y⟩ + ||y
Rollout 1
since k depends on ||y||2, our goal is to find y and k such that this equation
Rollout 3
793 * 0.222 ��0.697433.\n\n
Rollout 9
square is 80 square units, and I need to find the length of the altitude to EM in
Rollout 7
+A x=79*(P +x).\n\nTherefore,\n\n76P +A x=7
Rollout 4
8%)2 + (0.8%)2 )\n\nWait, actually, no. The formula involves relative
Rollout 8
5 of them in a row. We need to find the probability that all red cards laid out are adjacent
Rollout 7
. But the original scores can’t be98, since they need to be between60-6
Rollout 1
\) is \( d \), we need to show there exists a point \( y \in \math
Rollout 8
red cards in 5 positions). Similarly all green: 5!. So total 2*5!
Rollout 1
is a function from H to R, and we need f(x) = k for all x ∈ S
Top 16 Negative Activations
Rollout 1
bert space, even a countable set can be dense.\n\nAlternatively, we can use a mapping where y
Rollout 1
of arbitrary cardinality, but perhaps construction requires transfinite induction.\n\nAlternatively, if we realize that the set
Rollout 2
3, and 10 is 2*5, so 10 and 125
Rollout 1
are weakly compact.\n\nBut given that spheres are closed and convex. Wait, the sphere { y :
Rollout 1
since H is infinite-dimensional, it is a complete metric space. However, the intersection might still be empty
Rollout 1
right. The sphere is not convex. So perhaps closed balls would be convex, but spheres aren't.
Rollout 1
averaging may not work. But perhaps we can use completeness of the Hilbert space.\n\nAlternatively, if H
Rollout 2
(3) * (-1)^{997} mod8.\n\nFirst, compute (-1)^
Rollout 8
number is 10P5 = 10! / (10-5)! =
Rollout 1
is non-empty. So, since each sphere is closed and convex, and in a Hilbert space,
Rollout 1
ogonality condition comes automatically from the equidistant condition. So that's interesting. Therefore, if
Rollout 0
that \(a\) and \(b\) are coprime factors of 20!. So they form
Rollout 8
P5 = 10! / (10-5)! = 10 * 9
Rollout 2
by8. Since 5 and8 are coprime, m must be ��0 mod8
Rollout 1
of completing the system iterating over the points by transfinite induction. (Dragon a bit here.)\n\nAlternatively,
Rollout 0
), where \(k\) is the number of distinct prime factors of 20!.\n\nBut wait,

DOWN_PROJ

Top 16 Positive Activations
Rollout 9
square, then the sides are AI, IM, ME, and EA. Wait, maybe that's a
Rollout 5
area. \n\nFor a line segment parallel to the bases that divides the trapezoid into two regions
Rollout 2
Hmm, okay, let's unpack this step by step.\n\nFirst, the problem is asking for the remainder
Rollout 3
for approximation.\n\nWait, given that no temperature is given, perhaps they just want the formula, but I
Rollout 9
, which is 10; the bases are parallel sides? But wait the two lines at x=
Rollout 3
room temperature?\n\nBut surely, since no temperature is given, we must assume standard temperature, which is
Rollout 9
simpler way is to compute the area of the trapezoid formed between these two lines, from x
Rollout 5
Wait, let me check the calculations again step by step. Probably, I made an algebraic mistake.\n\n
Rollout 1
't state anything on how big the set S is. Given that in an infinite-dimensional Hilbert space,
Rollout 2
9 -1). Therefore, the product is ��_{k=1}^{999}
Rollout 0
say that a and b are coprime with a*b=20!, then a and b must
Rollout 2
product from k=1 to999 are even; for example, 9 is odd (mod
Rollout 8
* 6. Alternatively, that's 30240. But let me verify that.
Rollout 5
that right?\n\nWait, let me check the calculations again step by step. Probably, I made an algebra
Rollout 9
, IM horizontal, ME vertical downwards, and EA horizontal back to the start? Wait, now I'm
Rollout 3
8 m/s. So this checks out.\n\nSo all steps are correct.\n\nThus, conclusion: at
Top 16 Negative Activations
Rollout 1
such sets have certain properties.\n\nAlternatively, I might need to use the parallelogram law. Let me
Rollout 6
each encoding a bit of α_n. However, then S_L would have p(n) strings for each
Rollout 3
mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in the heavy gas radon (Rn
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a
Rollout 6
_L would contain all accepting paths, which might not be helpful.\n\nAlternatively, recall that P_angel is
Rollout 6
. [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these problems
Rollout 8
that's the number when each position can be red or green. However, since she's drawing without replacement

Layer 22

GATE_PROJ

Top 16 Positive Activations
Rollout 1
$ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show
Rollout 0
**\n\boxed{128}\n\n<|im_start|>answer\nGiven a rational number between 0 and 1
Rollout 2
109 in both cases. So The answer is109. Huh.\n\nBut wait,
Rollout 6
a sparse oracle, which is standard.\n\n<|im_start|>answer\n**Problem 1:**\n\nTo construct the sparse
Rollout 4
0.88\%}.\n\n<|im_start|>answer\nTo find the percentage error in the calculation of the
Rollout 0
128 such numbers. Therefore, The answer is 128, corresponding to \boxed{
Rollout 9
EM. Let me try to visualize this first. \n\nLet me draw the square AIME. Since it
Rollout 6
$L$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work
Rollout 0
is 128. Hence, the answer is 128?\n\nBut to confirm this,
Rollout 7
6}\n\n(b) No solution exists.\n\n<|im_start|>answer\nGiven a mathematics test with \( N \) participants
Rollout 7
-promoted $47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to
Rollout 5
is 181. Therefore, the answer is 181. \n\nWait, but let
Rollout 8
**\n\boxed{157}\n\n<|im_start|>answer\nKathy has 5 red cards and 5
Rollout 6
if the oracle says yes.\n\nTherefore, the answer is constructing S as the union of the tagged sets,
Rollout 1
}}" or as per problem statement.\n\n**[Final Answer]**\n\boxed{\text{Such a point
Rollout 2
**\n\boxed{109}\n\n<|im_start|>answer\nTo find the remainder when \(9 \times
Top 16 Negative Activations
Rollout 3
811.04. Divide by π*0.222≈0.69
Rollout 3
.314*298/(π*0.028)) �� sqrt(
Rollout 3
.314*298/(π*0.222)) �� sqrt(
Rollout 9
: y= (-5/25)x +50/25= (-1/5)x
Rollout 3
as earlier for radon)\n\nDenominator: pi*0.028= ~0.0
Rollout 3
,953.6; divided by π*0.222 �� 0.
Rollout 3
3 K:\n\n66.512 *293= compute 66.51
Rollout 7
new average 59. 54 +5=59. Correct.\n\nSo N=1
Rollout 3
19,811.04 /0.697433≈28
Rollout 4
Multiply by Δa:0.36 *10= 3.6 pF\n\nSecond
Rollout 7
76, as their original was71 +5) and x repeaters with average A+5
Rollout 3
in calculation:\n\n8 * R * T/(π*M): all three terms are three sig figs minimum
Rollout 3
0.576\n\nDenominator: π * M = π * 0.222
Rollout 4
energy:\n\nE_nominal=0.5 *1200e-12 *25
Rollout 7
86, new total=186 +5*3=201. Total promoted score
Rollout 7
.\n\nTherefore:\n\n(N/12)(A -54) = 2N/3\n\nDiv

UP_PROJ

Top 16 Positive Activations
Rollout 4
1/C_eq = 1/C1 + 1/C2. So, I can write that as
Rollout 6
sum over i of |S_i^{=n}|, which is at most k*p(n). Since
Rollout 4
get maximum energy, since energy is (1/2)*C_eq*V2, maximum energy would
Rollout 4
energy. The energy stored is (1/2) * C_eq * V^2. So,
Rollout 6
1 to k of |S_i^{=n}|. Since each S_i has at most p(n
Rollout 7
, total score becomes 66N + 5N = 71N. But also,
Rollout 9
, and their common area is 80 square units. Since the square itself is 10x
Rollout 4
in a capacitor is (1/2) * C_eq * V2. Since the question is about
Rollout 4
energy, since energy is (1/2)*C_eq*V2, maximum energy would be when
Rollout 4
.\n\nNow, since the energy is proportional to C_eq * V2, the relative error in energy will
Rollout 4
in this case, energy E is (1/2) * C_eq * V^2, so
Rollout 4
sqrt(3.62 +2.42)=sqrt(12.96+5
Rollout 4
The energy stored is (1/2) * C_eq * V^2. So, the energy
Rollout 7
total:79*(P + x)+47*(R -x)=71N.\n\nSame
Rollout 0
b = 20!\) and \(\gcd(a, b) = 1\), it
Rollout 4
the energy. The energy stored is (1/2) * C_eq * V^2. So
Top 16 Negative Activations
Rollout 2
is \boxed{109}\n\n**Final Answer**\n\boxed{109}\n\n<|im_start|>answer
Rollout 4
is approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of the energy
Rollout 3
{169} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon gas is
Rollout 5
81 is the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n<|im_start|>answer
Rollout 9
being h=25.\n\nPerfect.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>answer\n
Rollout 8
26= 157.\n\n**Final Answer**\n\boxed{157}\n\n<|im_start|>answer
Rollout 7
, part(b) has no solutions.\n\n**Final Answer**\n\n(a) \boxed{12}, \
Rollout 8
6= 157.\n\n**Final Answer**\n\boxed{157}\n\n<|im_start|>answer\n
Rollout 5
Wait, let me check the calculations again step by step. Probably, I made an algebraic mistake.\n\n
Rollout 2
\boxed{109}\n\n**Final Answer**\n\boxed{109}\n\n<|im_start|>answer\n
Rollout 3
{169} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon gas is
Rollout 4
approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of the energy stored
Rollout 3
169} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon gas is \
Rollout 0
-1}=128.\n\n**Final Answer**\n\boxed{128}\n\n<|im_start|>answer\n
Rollout 3
23, so if need to answer, in the absence of the temperature in the problem, I should
Rollout 0
8-1}=128.\n\n**Final Answer**\n\boxed{128}\n\n<|im_start|>answer

DOWN_PROJ

Top 16 Positive Activations
Rollout 7
71N.\n\nWhich simplifies as we did before to x=N/12.\n\nHowever, after
Rollout 7
-promoted $47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to
Rollout 5
the two bases. Wait, I might need to derive that. \n\nSuppose the original trapez
Rollout 2
since 1000 is 8 * 125 and 8 and 12
Rollout 7
)=71N.\n\nWhich simplifies as we did before to x=N/12.\n\nHowever,
Rollout 5
100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve
Rollout 9
0. Here's the reasoning: If h is small (h<10), the triangle is entirely
Rollout 9
=(10 - h,5). Because we found earlier that y=5. Even if h>
Rollout 0
, ordering matters. Therefore, for the purpose of our count, pairing where 2 and 3 are
Rollout 5
. Let me ensure that there were no algebraic mistakes. So the steps were:\n\n1. Found bases
Rollout 8
. Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We
Rollout 9
triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry problem
Rollout 3
in education is to round 168.5 up to 169. So probably,
Rollout 5
^2/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I
Rollout 1
$ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show
Rollout 2
to look at the problem modulo 8 and modulo 125 separately, since 100
Top 16 Negative Activations
Rollout 8
the number of sequences is the number of injective functions from 5 positions to 10 cards,
Rollout 4
/C_eq = 1/C1 + 1/C2. So, I can write that as C
Rollout 4
1/C_eq = 1/C1 + 1/C2. So, I can write that as
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a
Rollout 6
, so |S_L^{=m}| ≤1, which is certainly sparse. Thus, S_L
Rollout 6
number of strings for n would be p(n) �� p(2^m) = 2
Rollout 1
) for all x ∈ F has to be non-empty.\n\n2. By the completeness of H and the

Layer 23

GATE_PROJ

Top 16 Positive Activations
Rollout 4
in C_eq: sqrt( ( (10*3000^2)/(200
Rollout 4
5 -10*3000 +10*15=6,000
Rollout 4
\right)^2 + \left( 2 \cdot \frac{\Delta V}{V} \
Rollout 3
14 * 298 / (π * 0.222)) Hmm, okay
Rollout 7
x +59*(N/3) -59x =71N\n\nCalculate the terms
Rollout 4
03605)^2 + (2*0.004 )^2 )\n\nCompute
Rollout 7
:\n\n75*(2N/3) +75x +59*(N/3)
Rollout 4
/(5000)^2= (4*10^6)/(25*10
Rollout 4
2 * Δb )^2 = ( (2000^2)/(5000
Rollout 4
relative error in C_eq)^2 + (2 * relative error in V)^2 ) because the voltage
Rollout 3
222)) �� sqrt( (8*8.314*298)/(
Rollout 7
:\n\n79*(2N/3) +79x +47*(N/3)
Rollout 3
.\n\nAt 300 K: 8*8.314=66.5
Rollout 4
2000*15 -10*3000 +10*15
Rollout 4
sqrt(0.362 + (2*0.4)^2 )=sqrt(0
Rollout 4
}}} \right)^2 + \left( 2 \cdot \frac{\Delta V}{V}
Top 16 Negative Activations
Rollout 2
2^3 = 8 �� 0 mod 8, and 2^k for
Rollout 2
with remainders, especially modulo 1000, it often helps to look at the problem modulo
Rollout 2
0. But891 mod1000* (-1)= -891 mod1
Rollout 2
2^k for k ≥3 is 0 mod8. Thus, for k ≥3,
Rollout 2
\n\n2^3 = 0 mod 8\n\n2^4 = 0 mod 8\n\n
Rollout 8
+1=31. So 31. Denominator: 30240 �
Rollout 2
00. \n\nModulo 1000 only cares about the last three digits of the product
Rollout 2
We need to compute P mod 1000.\n\nNow, let's approach this modulo 8
Rollout 2
109 mod125: yes,109; 109 mod8
Rollout 2
8\n\n2^4 = 0 mod 8\n\nWait, 2^3 = 8
Rollout 5
0 = 181.25, greatest integer not exceeding 181.25
Rollout 2
can figure out the product modulo 1000, we can find the remainder. However, calculating
Rollout 5
1. Therefore, final answer is 181. \n\nI think that's correct. Let me
Rollout 2
≥3, 2^k �� 0 mod8, so (2^k -1
Rollout 2
then x≡109 mod1000. But wait, hold on. Wait,
Rollout 2
But we need the result modulo 1000. \n\nModulo 1000 only

UP_PROJ

Top 16 Positive Activations
Rollout 5
regions of equal area. \n\nFor a line segment parallel to the bases that divides the trapezoid
Rollout 1
x - y) x ∈ S, forms an orthonormal system. This means:\n\n• For each
Rollout 3
.\n\n66.512 * 273 �� 66.512
Rollout 1
Zorn's lemma, to find a maximal orthonormal system. But howdoes that help.\n\nHmm
Rollout 2
. For k>=3:2^k mod8. But 2^3=8≡0
Rollout 2
:\n\n9*99=891 mod1000\n\n891*99
Rollout 2
:\n\n(5m +5)≡5 mod8\n\nSubtract 5:\n\n5m≡0
Rollout 1
x ∈ S, forms an orthonormal system. This means:\n\n• For each x ∈ S,
Rollout 1
radius d / sqrt(2) centered at each point.\n\nBut in R^3, let's compute
Rollout 2
^k -1) �� -1 mod8.\n\nTherefore, for k=1: (2
Rollout 3
the mean (average) speed derived from the Maxwell-Boltzmann distribution:\n\n\[\nv = \
Rollout 9
10] [ (1/5 x +8 +1/5 x -2 ) ] dx
Rollout 3
=8*8.314*273=8*8.314=
Rollout 2
125\n\nx ��5 mod8\n\nLet x =125m +10
Rollout 5
25H.\n\nWe need to find a line parallel to the bases that splits the area into two equal
Rollout 2
125 is9 *99 * (-1). So, 9*99=8
Top 16 Negative Activations
Rollout 8
26 are coprime, so yeah, simplified.\n\nTherefore, the probability is 31/
Rollout 0
and 1. When written in lowest terms, numerator a and denominator b must satisfy that. However,
Rollout 7
56*8=448. Total of promoted-from-repeaters(original avg A)x=2
Rollout 5
the ratio of the bases is the same as the ratio of the heights. Wait, but these are not
Rollout 7
1$ and that of the repeaters $56$ . \nHowever, due to an
Rollout 3
inputs: 8 (exact)( since formula has 8, which is exact), R=8.
Rollout 6
related to n, but over all lengths, the total number for each length would be polynomial.\n\nWait,
Rollout 7
: their total original=124, new total=124 +5*2=1
Rollout 8
(5,5−k).\n\nThen we totaled 1200 +2400 +
Rollout 0
\(a \neq b\). Hence, total coprime pairs are \(2^{8}\
Rollout 7
): total score original=186, new total=186 +5*3=2
Rollout 7
0-64, sum /x=A must be possible?\n\nLet x=N/12. For
Rollout 5
+ b)/2 * h]. Then, simplifying, (a + x)k = (a
Rollout 2
given that our calculation says 109, which is not zero, but 109 and
Rollout 7
8*(N/12).\n\nBut sum must be integer. x is N/12.
Rollout 9
be entirely within the square. Thus, the area in common would be the area of the triangle which is

DOWN_PROJ

Top 16 Positive Activations
Rollout 0
k is primes. But, since prime factors with exponents: Wait, but each prime is either included
Rollout 0
factorization of 20!.\n\nTo compute the exponents of primes in 20!:\n\nFor
Rollout 0
not a square (unless for example primes with even exponents), it depends. As with N! for
Rollout 0
that's not correct. Each must have all primes exponents from 20! For example, if
Rollout 0
For each prime p | N, assign all the exponents in p to a or to b, but
Rollout 0
, actually 20! has prime factors with exponents. For example, 2^18
Rollout 0
is not a square (due to the primes with exponents as discussed earlier), there are no pairs where
Rollout 0
a = product_{p in S} p^{e_p}, and b = 20! /
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \
Top 16 Negative Activations
Rollout 1
cal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there exists
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this math
Rollout 6
. [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these problems
Rollout 8
m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to find
Rollout 9
triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry problem
Rollout 0
$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 3
) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular speed
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this
Rollout 2
1000. Since997 is odd, that's891*(-1)≡
Rollout 5
100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve
Rollout 8
$m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to
Rollout 2
97 is odd.\n\nTherefore, total product mod8 is 1*3*(-1) = -
Rollout 2
1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here:
Rollout 2
of -1 mod8.\n\nSo the product mod8 is (1) * (3) * (-

Layer 24

GATE_PROJ

Top 16 Positive Activations
Rollout 3
Pressure), which is 0°C and 1 atm. However, sometimes people use 25°C
Rollout 2
\n\nk=2:10^2 -1=99 mod125\n\nk=
Rollout 0
20/8) + floor(20/16) =10+5+2
Rollout 2
\n\nk=4:10^4 -1=9999≡(10^
Rollout 0
20/2) + floor(20/4) + floor(20/8)
Rollout 3
STP is 1 bar instead of 1 atm, but regardless, the formula for speed only depends
Rollout 3
73.15 K) and 1 atm. In physics, sometimes it's 20
Rollout 2
\n\nk=3:10^3 -1=999≡-1 mod12
Rollout 2
;\n\nk=2,10^2 -1=99\n\nk=3, so1
Rollout 3
8 K.\n\nCompute v_avg = sqrt(8*R*T/(π*M))\n\nLet's calculate the numerator
Rollout 0
20/4) + floor(20/8) + floor(20/16
Rollout 0
\n\nFor prime 3: floor(20/3) + floor(20/9)
Rollout 0
20/3) + floor(20/9) + floor(20/27
Rollout 3
273.15 K and 1 atm. At STP, the molar volume is
Rollout 0
20/9) + floor(20/27)=6 +2 +0=8
Rollout 0
20/7) + floor(20/49)=2 +0=2\n\nPr
Top 16 Negative Activations
Rollout 5
oid into equal areas. I think the formula is that the length of this line is the root mean square
Rollout 3
maybe the term "mean molecular speed" is defined such that it doesn't require temperature because it's given
Rollout 5
two bases. Wait, I might need to derive that. \n\nSuppose the original trapezoid
Rollout 3
(3RT/M). So perhaps the problem wants the average speed. However, in some contexts, "
Rollout 5
100. \n\nFirst, let me recall some trapezoid properties. In a trape
Rollout 3
K.\n\nAlternatively, maybe they expect us to state that the temperature is required? But no, the problem
Rollout 3
, maybe the term "mean molecular speed" is defined such that it doesn't require temperature because it's
Rollout 5
creates a certain area ratio. Let me also recall that the area of a trapezoid is (
Rollout 0
But wait, an alternative approach: suppose we recognize that each coprime pair (a,b) with
Rollout 3
we need to state standard assumptions.\n\nAlternatively, maybe the term "mean molecular speed" is defined such that
Rollout 5
heights squared is involved? Hmm, maybe I need to use similar figures. If the trapezoid
Rollout 4
, multiplied by 100%, I think. But first, I should probably find the relative errors
Rollout 9
assistant\n<|im_start|>think\nOkay, so I need to solve this geometry problem where there's a square named
Rollout 3
it's sqrt(3RT/M). So perhaps the problem wants the average speed. However, in some
Rollout 5
, no, let me check. \n\nI remember that for areas divided by a line parallel to the bases
Rollout 3
the term "mean molecular speed" is defined such that it doesn't require temperature because it's given in

UP_PROJ

Top 16 Positive Activations
Rollout 2
2^k -1) �� 0 mod 8, which would make the entire product congr
Rollout 9
the triangle GEM would be a large triangle with base EM (the right edge of the square) and
Rollout 2
3, 2^k �� 0 mod8, so (2^k -1)
Rollout 0
b\).\n\nNow, how do I count such pairs?\n\nAlternatively, in number theory, this is equivalent
Rollout 9
has two equal sides (GE and GM) with base EM.\n\nSo let's calculate coordinates.\n\nGiven:\n\n
Rollout 0
number of coprime pairs \((a, b)\) where \(a \times b =
Rollout 9
5) where h>10. Then the base EM is vertical from (10,0)
Rollout 2
≥3, 10^k≡0 mod125. Therefore, (10^
Rollout 0
number of coprime pairs \((a, b)\) with \(a < b\) is \
Rollout 2
} -1). We need to compute P mod 1000.\n\nNow, let's approach
Rollout 9
The area of this triangle can be calculated. The base EM is same as base from (10,
Rollout 2
x in:\n\nx �� 109 mod125\n\nx ��5 mod8
Rollout 2
\[\n x \equiv 5 \mod 8 \quad \text{and} \
Rollout 2
9=891. 891 mod125=16, then 1
Rollout 0
is equivalent to: how many pairs (a, b) are there with gcd(a, b)=1
Rollout 2
2^k for k ≥3 is 0 mod8. Thus, for k ≥3,
Top 16 Negative Activations
Rollout 5
2 = 75 ± 5*sqrt(725)\n\nSince y = x +
Rollout 3
298 K), their average speeds can be in the hundreds of m/s. For example, for
Rollout 5
(18125) which is exactly what we have. Thus, that is consistent.\n\nTherefore
Rollout 5
* 725) = 5 * sqrt(725). Is that right?\n\nWait
Rollout 4
1 * C2) / (C1 + C2). Then, the energy stored in a capacitor
Rollout 1
any two vectors from the centroid is a constant negative value. Wait, perhaps scaling them appropriately could make them
Rollout 3
no, temperature must be in Kelvin in the formulae.\n\nHence, given the confusion, perhaps the
Rollout 7
, due to an error in the wording of a question, all scores are increased by $5$
Rollout 3
compute the speed? Maybe they just want the formula expressed in terms of molar mass and R? No
Rollout 5
trapezoid into two regions with areas in the ratio 2:3. \n\nWait a sec
Rollout 4
000^2)/(5000^2) * 15 )^2\n\n
Rollout 5
*725) = 10*sqrt(725)\n\nSo y = [1
Rollout 2
when doing straight mod1000 I also arrived at109, which is the same final
Rollout 5
:\n\ny = 75 + 5*sqrt(725)\n\nThus, x +
Rollout 3
can use a different formula; sometimes these speeds are expressed in terms of a general formula with molar mass
Rollout 7
: "due to an error in the wording of a question, all scores are increased by 5.

DOWN_PROJ

Top 16 Positive Activations
Rollout 2
0 mod8\n\nThus, 5m≡0 mod8. So 5m is divisible by
Rollout 7
, but reading the problem again: "due to an error in the wording of a question, all scores
Rollout 0
\) are coprime factors of 20!. So they form a coprime pair whose product
Rollout 3
required? But no, the problem says "compute the mean molecular speed," so they must want a numerical
Rollout 2
8\n\nSubtract 5:\n\n5m≡0 mod8\n\nThus, 5m≡0
Rollout 0
are related to the divisors of 20! where each pair consists of two coprime numbers
Rollout 4
this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 9
, the apex is G. And the altitude to EM is the perpendicular distance from G to EM. Since
Rollout 7
's read the original problem again:\n\n"due to an error in the wording of a question, all scores
Rollout 5
/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to
Rollout 3
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 3
6.512 * 300 = 19,953.6 minus
Rollout 4
. Which does the question expect? The question says "percentage error in the calculation of the energy stored in
Rollout 0
$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how
Top 16 Negative Activations
Rollout 8
they can be simplified. Since 31 is prime. 126 divides by 2,
Rollout 0
and \(b\) are coprime, each prime factor in 20! must go entirely to
Rollout 5
x + 9375\n\nSubtract 125x + 9375
Rollout 9
to the start? Wait, now I'm confused. Wait squares have four sides. Let me actually assign
Rollout 9
+ y^2\n\nSubtract y^2 from both sides:\n\n0 = 100 -
Rollout 0
256\), but each pair is counted twice except when \(a = b\). But since
Rollout 5
2b + 150; subtract 2b from both sides: b + 7
Rollout 9
that would be the vertical side. Wait, no, in that labeling, if AIME is the square
Rollout 5
5x + 9375\n\nSubtract 125x + 937
Rollout 5
0 = 125y\n\nBring all terms to left side:\n\ny^2 - 2
Rollout 5
+ 150; subtract 2b from both sides: b + 75 =
Rollout 9
point G which is inside the square. Wait no, if the altitude is from G to EM, which
Rollout 0
), where \(k\) is the number of distinct prime factors of 20!.\n\nBut wait,
Rollout 0
128\), since each pair is counted twice. But actually, since we are considering ordered pairs
Rollout 9
triangle would lie entirely inside the square?\n\nWait no, if G is at (0,5), then
Rollout 7
ing 66P and 56R from both sides:\n\n5P = 10R

Layer 25

GATE_PROJ

Top 16 Positive Activations
Rollout 3
(3RT/M), where R is the gas constant, T is the temperature in Kelvin, and M
Rollout 3
8 K.\n\nCompute v_avg = sqrt(8*R*T/(π*M))\n\nLet's calculate the numerator
Rollout 3
the formula.\n\nv_avg = sqrt((8 * R * T) / (π * M))\n\nR
Rollout 1
the graph of a linear operator or something. Hmm.\n\nAnother approach. Assume H has an orthonormal
Rollout 3
the average speed is sqrt(8*8.314*298/(π*0
Rollout 3
's the average speed, which is sqrt(8RT/(πM)). The problem mentions "mean molecular
Rollout 3
speed formula is sqrt(3RT/M), where R is the gas constant, T is the temperature in
Rollout 3
RT/M) = sqrt(3*8.314*298/0.2
Rollout 0
is 2, which is even. Wait a second. Then primes 11, 13
Rollout 3
222 kg/mol.\n\nR is the gas constant, which is 8.314 J
Rollout 3
they must assume a temperature.\n\nWait, hold on. Maybe the problem is from a textbook where they assume
Rollout 3
/M) = sqrt(3*8.314*298/0.22
Rollout 1
is expressing this by defining various terms. Let me see.\n\nSuppose that all points in S lie in
Rollout 3
the formula expressed in terms of molar mass and R? No, the question is to compute the speed
Rollout 1
can’t assume it’s at the origin. Hmm.\n\nAlternatively, can we use the concept of a basis
Rollout 3
average speed is sqrt(8*8.314*298/(π*0.
Top 16 Negative Activations
Rollout 2
\boxed{109}\n\n**Final Answer**\n\boxed{109}\n\n<|im_start|>answer\n
Rollout 3
169} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon gas is \
Rollout 3
m/s, and rounded to three sig figs is 169 m/s. Thus, present
Rollout 0
-1}=128.\n\n**Final Answer**\n\boxed{128}\n\n<|im_start|>answer\n
Rollout 9
h=25.\n\nPerfect.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare
Rollout 1
as per problem statement.\n\n**[Final Answer]**\n\boxed{\text{Such a point } y
Rollout 7
no solution.\n\nBut the problem is translated probably expecting answer. Maybe my initial steps overlooked something.\n\nWait,
Rollout 8
6= 157.\n\n**Final Answer**\n\boxed{157}\n\n<|im_start|>answer\n
Rollout 0
}=128.\n\n**Final Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven a
Rollout 4
approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of the energy stored
Rollout 5
1 is the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n<|im_start|>answer\n
Rollout 5
the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n<|im_start|>answer\nLet the
Rollout 9
25.\n\nPerfect.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(A
Rollout 2
{109}\n\n**Final Answer**\n\boxed{109}\n\n<|im_start|>answer\nTo find
Rollout 3
168.5\n\nThus, answer is approximately 168.5 m/s.
Rollout 3
, \text{m/s}\).\n\n**Final Answer:**\n\[\n\boxed{169

UP_PROJ

Top 16 Positive Activations
Rollout 9
25.\n\nPerfect.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(A
Rollout 8
157.\n\n**Final Answer**\n\boxed{157}\n\n<|im_start|>answer\nKathy
Rollout 0
}=128.\n\n**Final Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven a
Rollout 5
the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n<|im_start|>answer\nLet the
Rollout 2
{109}\n\n**Final Answer**\n\boxed{109}\n\n<|im_start|>answer\nTo find
Rollout 2
0^k -1 is odd. Because even minus1 is odd. Therefore, every term in the
Rollout 1
problem statement.\n\n**[Final Answer]**\n\boxed{\text{Such a point } y \text
Rollout 7
no solutions.\n\n**Final Answer**\n\n(a) \boxed{12}, \boxed{24},
Rollout 7
These two groups: original promoted (scores70+ new) and original near repeaters (scores6
Rollout 1
all three. In R^3, the intersection of three spheres is likely a finite set or empty.
Rollout 3
69 m/s.\n\n**Final Answer**\n\boxed{170} m/s\n\nWait no
Rollout 2
and P≡5 mod8 gives rise to P≡109 mod1000. Wait
Rollout 3
this question is ambiguous. Nevertheless, likely answer \boxed{170} m/s. Alternatively,
Rollout 2
Chinese Remainder Theorem, got109 mod1000. However, the only way
Rollout 1
2.\n\nBut perhaps it's hard.\n\nWait, back to the equation:\n\nFor y to satisfy ||xi
Rollout 0
to \boxed{128}? Or wait, wait prime factors of 20! Are
Top 16 Negative Activations
Rollout 9
better way. Since each side is 10 units. So AI would be vertical, IM horizontal,
Rollout 9
a better way. Since each side is 10 units. So AI would be vertical, IM horizontal
Rollout 9
that's a better way. Since each side is 10 units. So AI would be vertical,
Rollout 0
20!\) and \(a < b\), so pairs (a,b) and (b
Rollout 8
example, how is the shuffling done? She shuffles the 10 cards (which are
Rollout 9
10,10). So these are straight lines. Let me find equations for GE and GM.\n\n
Rollout 1
an infinite set where each pair is separated by d, perhaps S must lie on such a quadratic manifold.
Rollout 9
So the square is AIME with sides 10 units each. Then we have an isosce
Rollout 9
it's a square, all sides are 10 units. Let's label the square with points A
Rollout 9
would be horizontal, splitting EM into two parts of 5 units each. But also, since EM is
Rollout 9
's a square, all sides are 10 units. Let's label the square with points A,
Rollout 2
999 nines. Each of these numbers is a string of 9s. The first
Rollout 9
's a better way. Since each side is 10 units. So AI would be vertical, IM
Rollout 9
the apex to the base splits the base into two equal parts and is perpendicular to the base. So since
Rollout 1
,\n\n||x - x'||2 = d2,\n\nwhich, as established, gives:\n\n||x||
Rollout 1
and the spheres have radius d / sqrt(2). Then, 2r = sqrt(2)

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
, temperature must be in Kelvin in the formulae.\n\nHence, given the confusion, perhaps the problem
Rollout 8
Wait, hold on. Wait, wait, no. If she is laying out 5 red cards,
Rollout 8
0. Wait, hold on. Wait, wait, no. If she is laying out 5 red
Rollout 0
to their exponents).. Wait, confusion arises here.\n\nWait, actually, we need to be precise.
Rollout 8
10. Wait, hold on. Wait, wait, no. If she is laying out
Rollout 0
of coprime ordered pairs, or is it?\n\nWait, but perhaps the counting is as follows:
Rollout 8
out of 10. Wait, hold on. Wait, wait, no. If she is laying
Rollout 7
slightly ambiguous. Let me recheck the problem statement.\n\n“due to an error in the wording of a
Rollout 9
Or is the altitude segment inside the triangle? Wait. Wait, in an isosceles triangle,
Rollout 9
the altitude segment inside the triangle? Wait. Wait, in an isosceles triangle, the altitude
Rollout 3
's given in terms of some standard conditions. Hmm.\n\nWait, in kinetic theory, mean speed depends on
Rollout 4
2 +something)%? Wait, no—wait:\n\nWait, the C_eq error is 0.
Rollout 6
�� m? Unlikely. For example, m is dominated by n^k, so for sufficiently
Rollout 8
confusing.\n\nWait, maybe not. Let's clarify.\n\nTotal number of ordered sequences of 5 distinct cards
Rollout 8
with two cards.\n\nWait, hold on. Wait, the original problem defines "all the red cards laid
Rollout 0
. But, since prime factors with exponents: Wait, but each prime is either included in a or
Top 16 Negative Activations
Rollout 3
8*R*T)/(π*M)), R is in J/(mol·K), which is same as (
Rollout 3
/(mol·K), which is same as (kg·m2)/(s2·mol·K
Rollout 8
126 is 2* 63=2*7*9. 31
Rollout 8
also not work. So we need to calculate the number of such "happy" sequences divided by the total
Rollout 8
/ 10P5\n\nAlternatively, compute numerator and denominator.\n\nFirst, compute denominator: 1
Rollout 3
, which is 8.314 J/(mol·K).\n\nAssuming standard temperature. Since
Rollout 3
constant, which is 8.314 J/(mol·K).\n\nAssuming standard temperature.
Rollout 0
what about multiplicity. For example, 1 and 720 were co-prime (a
Rollout 7
original promotion. Therefore, N must be a common multiple of 3 and 12, that is
Rollout 4
.000076996. sqrt(0.000076
Rollout 3
), which is same as (kg·m2)/(s2·mol·K). Then,
Rollout 1
on the sphere's surface, and all the vectors from y to x are mutually orthogonal.\n\nBut wait,
Rollout 8
$\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers
Rollout 3
is sometimes given as 8.314 J/(mol·K), sometimes 8.3
Rollout 3
*R*T)/(π*M)), R is in J/(mol·K), which is same as (kg
Rollout 3
in denominator; thus overall (kg·m2)/(s2·mol) divided by (kg/mol

Layer 26

GATE_PROJ

Top 16 Positive Activations
Rollout 3
Let's see, 168^2 = 28,224. 1
Rollout 3
:\n\n70*66.512=4,655.84\n\n3
Rollout 5
500) = sqrt(100*725) = 10*sqrt
Rollout 4
75 �� Compute 4975*1190=4975*(
Rollout 4
4975*1190=4975*(1200-
Rollout 4
,000 and 2000*15=30,000 and
Rollout 4
5025 * 1200 = 6,030,000
Rollout 8
's see: 240×30 =7200, 7440
Rollout 3
4\n\n3*66.512=199.536\n\nTotal=
Rollout 3
66.48*300=19,944 – 66
Rollout 5
= sqrt(100*725) = 10*sqrt(725
Rollout 5
8125) = sqrt(25 * 725) = 5 * sqrt
Rollout 4
015 = Compute 2000*3000=6,000
Rollout 3
*66.512:\n\n70*66.512=4,6
Rollout 3
66.512 =\n\n200*66.512=13,
Rollout 3
3= compute 66.512*290=19,288
Top 16 Negative Activations
Rollout 0
, numerator a and denominator b must satisfy that. However, a different way to phrase this is: a
Rollout 7
. However, the problem states N<40. 36 is less than 40,
Rollout 9
would be outside the square.\n\nBut the problem states that the area common to triangle GEM and square A
Rollout 7
.\n\nSame result, which is impossible. Therefore, the conditions given in part(b) areimpossible.\n\n
Rollout 0
be resolved.\n\nWait, perhaps that's the confusion. The original number is some fraction as part of between
Rollout 3
the rms speed. If they wanted the rms speed, but the question says "mean".\n\nAlternatively, maybe
Rollout 3
" speed, it's probably the average speed formula. But maybe I should check. If I can't
Rollout 7
mistake in assumptions, perhaps the problem in part(b) is impossible? But no, the question is "
Rollout 0
0 factorial. Wait, that needs to be resolved.\n\nWait, perhaps that's the confusion. The original
Rollout 7
. The pass mark is still 65. Therefore, after the increase, to be promoted, you
Rollout 7
so maybe there are no solutions? But likely, the problem expects there to be solutions. So I must
Rollout 9
, which in this case is the vertical side. However, an isosceles triangle with base EM
Rollout 3
/M). So perhaps the problem wants the average speed. However, in some contexts, "mean molecular speed
Rollout 2
09 mod125.\n\nHence, the solution is x≡109 mod10
Rollout 7
=24 is impossible.\n\nTherefore, part(b) has no solutions.\n\n**Final Answer**\n\n(a)
Rollout 3
but I might have to verify which one is the actual mean speed. Let me think.\n\nMean speed,

UP_PROJ

Top 16 Positive Activations
Rollout 9
triangle GEM and square AIME is 80. Since the square is from (0,0
Rollout 5
(H - h) = 62.5H.\n\nSo we have two equations:\n\n1.
Rollout 5
x)/2 * h = 62.5H. The remaining trapezoid has bases
Rollout 9
\) and square \(AIME\) is 80 square units. We need to find the length of
Rollout 9
to both the triangle and the square is 80 square units, and I need to find the length
Rollout 4
(C1 + C2)) ] + dC2 [ (C1 + C2 - C
Rollout 8
total number of possible sequences is 30240.\n\nNow, we need to count how many
Rollout 4
\frac{\Delta C_{\text{eq}}}{C_{\text{eq}}} \right
Rollout 3
gas is \boxed{170} m/s.\n\nWait, after saying 168.
Rollout 9
triangle GEM and square AIME is 80 square units. The square area is 10
Rollout 9
inside the square (the common area is 80), and the square's right edge is EM,
Rollout 1
x⟩ + ⟨y, y⟩ - d2 / 2.\n\nBut this is a system of
Rollout 4
(C1 + C2)) ] + dC2 [ C1 / (C2 (C
Rollout 4
C_eq was 4.327 pF, C_eq is 1200 p
Rollout 4
1/(C1 + C2)] + dC2 [1/C2 - 1/(C
Rollout 5
2b + 150)/2 * h/2 = (b + 75)
Top 16 Negative Activations
Rollout 3
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 4
this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage
Rollout 0
$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how
Rollout 9
$\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry
Rollout 2
1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here:
Rollout 5
/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to
Rollout 8
$m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this
Rollout 4
combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage error
Rollout 0
be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how many
Rollout 3
) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular speed
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this math
Rollout 5
100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve
Rollout 1
cal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there exists
Rollout 2
000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here: I

DOWN_PROJ

Top 16 Positive Activations
Rollout 4
990 * 2985= (2000 -10)*(30
Rollout 3
3 * 66.512 = (200 +70 +3)*6
Rollout 3
is 300 - 2, so subtract 66.512 * 2
Rollout 2
mod1000 is109. Correct.\n\nFour terms:9*99*9
Rollout 4
)=4975*1200 -4975*10=5,
Rollout 4
0036055 squared: Approximately (0.0036)^2 =
Rollout 2
09*999=109*(1000-1)=109
Rollout 3
sqrt(x)=168.5.\n\nSo sqrt(28,392.25
Rollout 3
4.\n\n168.5^2= (168 + 0.5)^2
Rollout 3
, which is 300 - 2, so subtract 66.512 *
Rollout 3
Let's see.\n\n168^2 = (170 - 2)^2 =
Rollout 2
. But let me check this process again.\n\nFirst modulus8:\n\nProduct=(10^1 -1
Rollout 7
1*24=1704. Okay, so totals adds up. But x=N/
Rollout 3
which is 300 - 2, so subtract 66.512 *
Rollout 8
7440 divided by 240. Let's see: 240×3
Rollout 3
2^2=26,244. So it's between 161 and
Top 16 Negative Activations
Rollout 3
on (Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate
Rollout 8
block and all the green cards are in a block. But since she's only laying out 5 cards
Rollout 0
^{}!$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to
Rollout 4
energy stored in this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to
Rollout 0
subset) and one unique b (multiplying the rest).\n\nSince each subset has a unique complement, the
Rollout 1
||2 - d2)/2.\n\nBut we don't know if all points in S have the same
Rollout 3
which is related to temperature and molar mass. Wait, but the question specifically says "mean molecular speed
Rollout 4
energy with C1 max and min, C2 max and min, and V max and min. But
Rollout 0
\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number, write it as a fraction
Rollout 1
\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$
Rollout 3
\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in the heavy gas
Rollout 4
\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values $C_{
Rollout 6
\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $L \subseteq
Rollout 7
\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants is $
Rollout 8
\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red cards and $
Rollout 9
\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides of length $

Layer 27

GATE_PROJ

Top 16 Positive Activations
Rollout 5
.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve this problem about a
Rollout 8
.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to find the probability that when
Rollout 7
$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this math problem about participants
Rollout 0
product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how many rational numbers between
Rollout 2
$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here: I need to find
Rollout 9
$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry problem where there's
Rollout 3
in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular speed of
Rollout 5
00$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve this
Rollout 4
of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage error in
Rollout 4
itors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage error in the calculation
Rollout 9
GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry problem where
Rollout 1
}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there exists a point y
Rollout 8
+ n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to find the
Rollout 1
{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there exists a
Rollout 6
me try to work through these problems one by one. They all seem related to sparse sets, oracles
Rollout 6
[/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these problems one
Top 16 Negative Activations
Rollout 0
count above). But 2^{3-1}= 4 correct. Hence, Therefore, the formula
Rollout 0
4, which is 2^{3-1}=4.\n\nSimilarly, if n is a square-free
Rollout 0
128. Alternatively, total number of subsets is 2^8=256, each
Rollout 9
would be (10 *10)/2 =50, so overlap area is 50
Rollout 2
^k mod8. But 2^3=8≡0 mod8, 2^k
Rollout 4
14.8058)/15 = 0.39/15 ��
Rollout 8
5*4*3 * 5*4= 60*20=120
Rollout 8
10*9*8*7*6≈7440 /30240
Rollout 4
, so total≈14,328 +477.8388≈1
Rollout 8
40= Let's check by dividing numerator and denominator by 240: 744
Rollout 0
20 is 4, which is 2^{3-1}=4.\n\nSimilarly, if n
Rollout 0
problem. It's 2^{8-1}=128. Thus the answer is 1
Rollout 0
. Since the total number is 2^8=256 (since each prime is either assigned
Rollout 8
0= Let's check by dividing numerator and denominator by 240: 7440
Rollout 4
(3.62 +2.42)=sqrt(12.96+5.
Rollout 0
, total number of subsets is 2^8=256, each ordered pairs. Taking half

UP_PROJ

Top 16 Positive Activations
Rollout 2
8, so 10^k mod 8 is 2^k mod 8.\n\nThus
Rollout 2
(10^k -1) mod 8 is (2^k -1) mod
Rollout 2
Since 10 �� 2 mod 8, so 10^k mod 8
Rollout 2
^k -1) �� -1 mod8.\n\nTherefore, for k=1: (2
Rollout 2
8.\n\nCompute 2^k mod 8:\n\nNotice that 2^1 = 2
Rollout 9
+ (10 - y)^2]\n\nSince GE = GM, then:\n\nsqrt[h^2 +
Rollout 2
0 -1)= -1 ��7 mod8\n\nk=4: same as k=3
Rollout 2
1) = 2 -1 =1 mod8\n\nk=2: (4 -1)=
Rollout 7
originally scored 60 would now have 65, so they would be promoted. So some repe
Rollout 2
is (2^k -1) mod 8.\n\nTherefore, P mod 8 is the product
Rollout 1
2) } is not convex. Oh, right. The sphere is not convex. So perhaps closed balls
Rollout 2
that 2^1 = 2 mod 8\n\n2^2 = 4 mod 8
Rollout 2
8\n\n2^3 = 0 mod 8\n\n2^4 = 0 mod 8
Rollout 9
with GE = GM.\n\nCalculating GE distance:\n\nGE is from G (10 - h, y
Rollout 2
mod 8 is 2^k mod 8.\n\nThus, each term (10^k
Rollout 1
Oh, right. The sphere is not convex. So perhaps closed balls would be convex, but spheres aren
Top 16 Negative Activations
Rollout 2
, then combine the results using the Chinese Remainder Theorem.\n\n1. **Modulo 8 Calculation
Rollout 0
the product \(a \times b = 20!\).\n\n1. **Prime Factorization of
Rollout 3
not specify the temperature, we assume a standard temperature of \(298 \, \text{K
Rollout 3
:**\n - Compute the numerator: \(8 \times 8.314 \times
Rollout 7
so \( N \) must be a multiple of 3.\n\n2. **After score increase:**\n
Rollout 9
0)\), \(M(10, 10)\), and \(E(10,
Rollout 4
for capacitors in series is given by:\n\n\[\nC_{\text{eq}} = \frac
Rollout 7
= 2R \) and \( N = 3R \), so \( N \) must
Rollout 4
energy stored in two capacitors connected in series, we start by determining the equivalent capacitance \( C_{
Rollout 2
+ 109\). Then:\n \[\n 125m + 1
Rollout 3
molecular speed of radon gas is \boxed{170} m/s.\n\nWait, after saying
Rollout 2
boxed{109}\n\n<|im_start|>answer\nTo find the remainder when \(9 \times 99
Rollout 4
^2 + 2.4^2} = \sqrt{12.96 +
Rollout 7
so \( N \) must be a multiple of 12. Possible values: 12,
Rollout 4
.96 + 5.76} = \sqrt{18.72} \
Rollout 8
, which is \(10 \times 9 \times 8 \times 7 \times

DOWN_PROJ

Top 16 Positive Activations
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $A
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of
Rollout 4
3015)/(2010 + 3015). Let's compute that
Rollout 9
] dx\n\nWait, wait subtract the lower from upper:\n\n= ��[0 to10]
Rollout 9
5)x +8.\n\nThus, calculate the area between these two lines from x=0 to x=
Rollout 2
: 9 *99 * (-1)^{997} mod125\n\nBecause
Rollout 9
2 +8=10.\n\nThus, area between these two lines is the area of trapez
Rollout 2
now we have 16 * (-1)^{997} mod125.\n\nSince
Top 16 Negative Activations
Rollout 7
the new average of the promoted is 75 and non-promoted is 59. The total
Rollout 2
. When k=1: 9 is odd. 10^k-1 in general is
Rollout 6
Problem 1:**\n\nTo construct the sparse set \( S \), we can combine the sparse sets \(
Rollout 6
classes. Let's start with the first one.\n\n**Problem 1:** Given k sparse sets S_
Rollout 7
10R\n\nDivide both sides by 5:\n\nP = 2R\n\nSo in
Rollout 7
, original score + 5 must be >=65? Wait, but if a student's original score
Rollout 0
terms, that means that a and b are coprime, hence exponents of primes in 2
Rollout 0
\((a, b)\) where \(a \times b = 20!\) and \(
Rollout 7
>=60 and <65. 98 not possible.\n\nN=24: x=
Rollout 0
and \(a\) and \(b\) are coprime positive integers. The product \(a \times
Rollout 5
. Cross-multiplying: 3(b + 25) = 2(b + 7
Rollout 7
x scored 64). So A=98 is impossible.\n\nTherefore, inconsistency. What does that
Rollout 7
192, but equations require294, impossible. Hence there's no solution. Thus part
Rollout 7
if a student's original score +5 >=65, that's original score >=60. Wait
Rollout 7
is about the value of calculatedA=98 (original part(b)) which is impossible as their original
Rollout 7
1, so their new average would be 76.\n\nBut in the problem statement, after the score

Layer 28

GATE_PROJ

Top 16 Positive Activations
Rollout 3
olar mass and R? No, the question is to compute the speed, so they must assume a temperature
Rollout 0
20!.\n\nHence, the question is equivalent to: how many pairs (a, b)
Rollout 1
subject.\n\nAlternatively, but, finally, the answer is asserting that such a y exists because he’s able
Rollout 3
that T is needed. But since the question is to compute, we proceed assuming 298 K
Rollout 3
confirm. But let's see, the question is asking for the "mean molecular speed v"—maybe the
Rollout 1
∈S} is orthonormal.\n\nWhich is equivalent to requiring that { x - y } is orthogonal
Rollout 3
61.4 m/s.\n\nAs an assistant, I need to perhaps suggest both possibilities, but likely
Rollout 8
"the probability that Kathy will be happy", which is dependent on the arrangement of colors. So if there
Rollout 8
But the way the problem is phrased: "all the red cards laid out are adjacent and all
Rollout 3
specify the temperature. Hmm. But since they said to compute it, maybe we need to state standard assumptions
Rollout 1
equations in H for y.\n\nNow, the set of solutions to a single equation ||x - y||
Rollout 0
is equal to 20!. So that equivalent to saying find all reduced fractions a/b where 0
Rollout 1
d2 / 2. Which can be considered as equations in H for y.\n\nNow, the set
Rollout 4
how they expect, 0.88% or 0.9%.\n\nBut given the original data
Rollout 1
x∈S} is orthonormal.\n\nWhich is equivalent to requiring that { x - y } is
Rollout 9
square AIME is 80, which is 80% of the square. Therefore, the
Top 16 Negative Activations
Rollout 7
:**\n - Total score: \( 71N = 75(P + x) +
Rollout 7
: \( 71N = 75(P + x) + 59(R - x
Rollout 7
: \( 71N = 79(P + x) + 47(R - x
Rollout 7
2N/3) + x ) +59*( (N/3) - x ) =
Rollout 7
75(P + x) + 59(R - x) \)\n - Substit
Rollout 3
8.314 \times 298 = 19,820.
Rollout 7
79(P + x) + 47(R - x) \)\n - Solving similarly
Rollout 4
{2000 \cdot 3000}{2000 + 30
Rollout 0
2^8 = 256\) coprime pairs \((a, b)\).\n\n3
Rollout 2
equiv 109 \mod 125\n \]\n\n3. **Combining Results
Rollout 7
47*(N/3 -x)=71N.\n\nLet's expand:\n\n79*(2
Rollout 2
times (-1) \equiv -891 \equiv -16 \equiv 109
Rollout 7
75(P + x) + 59(R - x) \)\n - Substituting
Rollout 2
equiv -891 \equiv -16 \equiv 109 \mod 12
Rollout 7
79(P + x) + 47(R - x) \)\n - Solving
Rollout 7
)+47(N/3 -x)=71N.\n\nWhich simplifies as we did before to

UP_PROJ

Top 16 Positive Activations
Rollout 4
000, since 5025*1000=5,025
Rollout 3
). Let's see, 168^2 = 28,224.
Rollout 9
(10 - y)^2]\n\nSquaring both sides:\n\nh^2 + y^2
Rollout 3
28,408.12-28,392.25=
Rollout 3
1.4, as 161^2=25,921;16
Rollout 4
Then, their product is 1990*2985. Let's compute that.\n\n
Rollout 4
25:\n\nFirst, 5025 * 1200 = 6,0
Rollout 5
sqrt[(752 + 1752)/2] = sqrt[(5625
Rollout 4
*C2)/(C1 + C2). To maximize this expression, ideally if C1 and C2
Rollout 5
me substitute h from the first into the second equation.\n\nSo:\n\n(x + 175)(H
Rollout 3
3= compute 66.512*290=19,288
Rollout 9
50, so overlap area is 50, as before. So how can we have overlapping area
Rollout 9
+ (10 - y)^2]\n\nSquaring both sides:\n\nh^2 + y^
Rollout 0
6 < b=45? 16 <45: yes.\n\n3. Assign 3
Rollout 8
7200= 7440. Total sequences 10P5=30
Rollout 4
6,030,000, since 5025*100
Top 16 Negative Activations
Rollout 9
,10), (0,10 -50/h), (0,50/h),
Rollout 7
) \boxed{12}, \boxed{24}, \boxed{36}\n\n(b)
Rollout 9
0), (0,10 -50/h), G's projection path, etc. Wait,
Rollout 9
0/h) and (0,10 -50/h), which is between (0, approximately
Rollout 9
/h) and K (0,10 -50/h).\n\nBut 10 -50
Rollout 4
004% error, but the voltage uncertainty was ±0.02 V on 5.
Rollout 9
0), (0,10 -50/h), (0,50/h), and back
Rollout 9
0), (0,10 -50/h), (0,50/h).\n\nBut actually
Rollout 1
2α - ||y||2 = c - d2/2.\n\nSo substituting back into the
Rollout 3
m/s.\n\nBut the initial calc uses T = 298 K:\n\n8.314
Rollout 9
0), (0,10 -50/h), (0,50/h). Connecting (
Rollout 2
5 is: 9 *99 * (-1)^{997} mod125
Rollout 2
9 mod125 and P≡5 mod8 gives rise to P≡109 mod1
Rollout 1
(||x||2 + ||x'||2 - d2)/2 - ⟨x + x', y
Rollout 1
+ (||t||2)/4 - (d2)/2.\n\nBut now this is getting more
Rollout 9
,10), (0,10 -50/h), G's projection path, etc.

DOWN_PROJ

Top 16 Positive Activations
Rollout 2
109 in both cases. So The answer is109. Huh.\n\nBut wait,
Rollout 8
.246.\n\nThus, the answer would be 31 +126=15
Rollout 0
-1}=128. Thus the answer is 128. Hence, the answer is
Rollout 6
conjectured to be false. Thus, the answer is P = P_bad-angel, and NP ≠
Rollout 0
20! is not just the product of primes, but the product of primes raised to various powers.
Rollout 6
is impossible in polynomial time.\n\nSo perhaps the answer is not straightforward, hence why reference to the P/p
Rollout 0
128 such numbers. Therefore, The answer is 128, corresponding to \boxed{
Rollout 8
For k=3:\n\nSame as k=2 but reversed. Because k=3 reds and
Rollout 7
R = 66P + 66R\n\nSubtracting 66P and
Rollout 9
the altitude h is25. So the answer is 25. Therefore box form \boxed{
Rollout 1
equation derived from orthogonale:\n\nFrom above, using orthogonality condition:\n\n||y||
Rollout 5
is 181. Therefore, final answer is 181. \n\nI think that's
Rollout 5
H - h) = 62.5H.\n\nSo we have two equations:\n\n1. (
Rollout 0
28 must be correct, hence The answer would be 128.\n\nBut before closing,
Rollout 5
is 181. Therefore, the answer is 181. \n\nWait, but let
Rollout 7
iven these constraintsthe answer to part(b) is that there is no possible value of N.\n\nBut
Top 16 Negative Activations
Rollout 1
2) d �� 1.414d. The mutual distance between centers is d.
Rollout 9
(10 -5)/(10 - (10 - h)) = 5/h.\n\nTherefore
Rollout 8
sequences must be either R^r G^{5−r} or G^{5−r} R
Rollout 7
wording is slightly ambiguous. Let me recheck the problem statement.\n\n“due to an error in the wording
Rollout 3
). So sqrt(8)=2.828. 475 /2.82
Rollout 3
they use 273 K, others 298 K. For example, if it's
Rollout 8
, we need to consider all possible partitions where there's at least one red and one green. Also,
Rollout 8
0.\n\nWait, but hold on. Let me check for k=3.\n\nWait, k=3
Rollout 4
3, so 2.774 * 3.1623e-3 �
Rollout 7
is P =2N/3. Let me check derived value.\n\nOriginal equations established from before:\n\nTotal
Rollout 1
2 = d2/2.\n\nRearranged:\n\n2⟨xi, y⟩ = ||xi
Rollout 7
1P +56(N - P) = 66N.\n\nSimplify:\n\n71P
Rollout 9
).\n\nDistance GM = sqrt[(10 - (10 - h))^2 + (10 -
Rollout 3
314=66.512; 273.15*66
Rollout 1
the various norms ||x|| are the same, maybe we can attempt to define y by some averaging process
Rollout 4
6,060,150 / 5025. \n\nDividing 6

Layer 29

GATE_PROJ

Top 16 Positive Activations
Rollout 8
BB... Wait, GGRR would actually be similar: P(5,2)*P(5,
Rollout 9
dx\n\nWait, wait subtract the lower from upper:\n\n= ��[0 to10] [
Rollout 4
. Maybe this will increase precision?\n\nWait, or rather:\n\nWait, before the coefficient for ΔC1
Rollout 1
lie on such a sphere. Wait, no. What we need is more along the lines of for each
Rollout 9
EA. Wait, maybe that's a better way. Since each side is 10 units. So
Rollout 4
002 *10, etc. Let's recalculate the derivatives values more precisely:\n\nStarting with
Rollout 4
But wait, this might be another approach. Let's see:\n\nIf I take the natural log of C
Rollout 8
-r))! ; wait, maybe it's better to think:\n\nWait, arranging r reds and
Rollout 8
positions and green cards to the green positions. Let me see.\n\nFor a specific color sequence, say RR
Rollout 4
, this might be another approach. Let's see:\n\nIf I take the natural log of C_eq =
Rollout 8
, sorry, k is the number of red cards. So for k=2, the color sequences are
Rollout 8
...R.\n\nBut actually, wait: If you have two blocks, you can have either R followed by
Rollout 9
(10,0). Wait, wait no: in the problem statement, the square is called A
Rollout 9
, if AIME is the square, then the sides are AI, IM, ME, and EA.
Rollout 8
...GR...R.\n\nBut actually, wait: If you have two blocks, you can have either R
Rollout 8
! ; wait, maybe it's better to think:\n\nWait, arranging r reds and 5−
Top 16 Negative Activations
Rollout 1
cal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there exists
Rollout 3
) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular speed
Rollout 2
000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here: I
Rollout 4
combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage error
Rollout 0
be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how many
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this math
Rollout 6
. [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these problems
Rollout 8
m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to find
Rollout 5
100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve
Rollout 4
this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 3
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 2
1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here:
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this
Rollout 9
triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry problem

UP_PROJ

Top 16 Positive Activations
Rollout 0
b\). So for each such coprime pair where \(a < b\), we have a
Rollout 0
and \(b\) are coprime, each prime factor in 20! must go entirely to
Rollout 0
!, where \(a\) is the product of the primes in the subset and \(b\) is the product
Rollout 0
subset of the prime factors of 20!, where \(a\) is the product of the primes in
Rollout 0
ime.\n\nHence equivalence between choices over subsets of primes.\n\nHence, since ordered coprimes,
Rollout 0
2^8=256 (since each prime is either assigned to a or to b,
Rollout 6
n, you can't encode solutions for all possible formulas of size n because SAT is NP-complete. Thus
Rollout 1
{ y + (d / sqrt(2})) e_x } where {e_x} is orthorn
Rollout 0
problem reduces to finding the number of coprime pairs \((a, b)\) where \(a
Rollout 7
is 71, and the average of those who failed (repeaters) is 56.
Rollout 0
applies; each such pair is obtained by assignment of primes to a or b, since a and b need
Rollout 0
But is there a simple relation between the subset of primes assigned to a and whether a < sqrt(N)?\n\n
Rollout 0
a\) or \(b\). So for each prime factor, we assign it to either \(a\)
Rollout 0
b\).\n\nNow, how do I count such pairs?\n\nAlternatively, in number theory, this is equivalent
Rollout 6
each m, there's at most one n (unless the encoding is overlapping). Hence, the number of
Rollout 0
, this is equivalent to counting the number of ways to write 20! as a product of two
Top 16 Negative Activations
Rollout 5
5 is 181.\n\n\[\n\boxed{181}\n\]<|im_end|>
Rollout 8
= 157\).\n\n\[\n\boxed{157}\n\]<|im_end|>
Rollout 9
answer is 25. Therefore box form \boxed{25}\n\nWait, that seems correct.\n\n
Rollout 8
6= 157.\n\n**Final Answer**\n\boxed{157}\n\n<|im_start|>answer\n
Rollout 0
answer is 128, corresponding to \boxed{128}? Or wait, wait
Rollout 1
as per problem statement.\n\n**[Final Answer]**\n\boxed{\text{Such a point } y
Rollout 8
157.\n\n**Final Answer**\n\boxed{157}\n\n<|im_start|>answer\nKathy
Rollout 8
26= 157.\n\n**Final Answer**\n\boxed{157}\n\n<|im_start|>answer
Rollout 2
\boxed{109}\n\n**Final Answer**\n\boxed{109}\n\n<|im_start|>answer\n
Rollout 5
the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n<|im_start|>answer\nLet the
Rollout 3
{169} m/s.\n\n<|im_start|>answer\nTo compute the mean molecular speed \( v \)
Rollout 3
boxed{169} m/s.\n\n<|im_start|>answer\nTo compute the mean molecular speed \( v \
Rollout 5
Answer**\n\boxed{181}\n\n<|im_start|>answer\nLet the shorter base be \( b \)
Rollout 2
109.\n\nTherefore, boxed answer is \boxed{109}\n\n**Final Answer**\n\
Rollout 7
.\n\nTherefore, part(b) has no solutions.\n\n**Final Answer**\n\n(a) \boxed{12
Rollout 4
{0.88\%}.\n\n<|im_start|>answer\nTo find the percentage error in the calculation of

DOWN_PROJ

Top 16 Positive Activations
Rollout 6
all accepting paths, which might not be helpful.\n\nAlternatively, recall that P_angel is equivalent to P
Rollout 3
answer as around 170 m/s. Alternatively, let's check some resource. Wait, can
Rollout 3
:\n\nWe did 161 m/s.\n\nAlternatively, textbook's choice. Given a chemistry book,
Rollout 1
D, i.e., tetrahedron, but even a triangle is 2D.\n\nWait,
Rollout 6
where i is a bit position in α_n, but then S_L contains these pairs (n,i)
Rollout 1
not necessarily have any kind of limit. Maybe.\n\nAlternatively, take y = average of all points in S
Rollout 3
.31, depending on the precision needed.\n\nAlternatively, maybe the original question assumes a specific temperature but
Rollout 6
s_n be the binary encoding of α_n. But α_n has length p(n), so s_n
Rollout 4
8%.\n\nAlternatively, for more precise fractional calculation.\n\nAlternatively, perhaps exact calculus on error, but since we
Rollout 3
. Alternatively, maybe exact problem has specific answer.\n\nAlternatively, maybe the problem uses R=8.3
Rollout 1
possibly uncountable, this isn't feasible.\n\nAlternatively, take a countable subset of S and construct
Rollout 6
the prior idea which leads to non-sparsity.\n\nAlternatively, define S_L as follows: For each α
Rollout 3
9} m/s. But need to specify.\n\nAlternatively, here's another angle: gas thermometric speeds
Rollout 1
n-dimensional space with n - 1 points, but since our space is infinite-dimensional, we can embed
Rollout 1
80 year old theorem on the subject.\n\nAlternatively, but, finally, the answer is asserting that
Rollout 8
all red cards in the layout are adjacent (but if there are red cards and green cards, then the
Top 16 Negative Activations
Rollout 3
Wait, but I might have to verify which one is the actual mean speed. Let me think.\n\nMean
Rollout 1
zero. So, the orthogonality condition comes automatically from the equidistant condition. So that
Rollout 9
50, as before. So how can we have overlapping area of 80? Since the square
Rollout 5
8125) which is exactly what we have. Thus, that is consistent.\n\nTherefore, x
Rollout 5
18125) which is exactly what we have. Thus, that is consistent.\n\nTherefore,
Rollout 5
(18125) which is exactly what we have. Thus, that is consistent.\n\nTherefore
Rollout 8
single block each.\n\nSo that makes sense. So how else can we model this?\n\nPerhaps treating the sequences
Rollout 5
x)\n\nSo the equation becomes:\n\n(x + 175)(x - 50)/(7
Rollout 4
the previous result.\n\nTherefore, combining this with the 2*0.004 (from the
Rollout 9
, so triangle is within. But then how come the overlapping area is 80 then? Because that
Rollout 8
, for color arrangements where all reds are together and all greens are together, regardless of the order (
Rollout 0
actually, we need to be precise. When we say that a and b are coprime with a
Rollout 4
F. C_eq=(2000*3000)/(2000+3
Rollout 7
Therefore, A=42 +56=98.\n\nBut that's impossible as A must be
Rollout 4
product over the sum, so C_eq = C1C2/(C1 + C2). The
Rollout 4
200. Yep, that's right. So, C_eq is 1200 p

Layer 30

GATE_PROJ

Top 16 Positive Activations
Rollout 9
base \(EM\), and the area common to triangle \(GEM\) and square \(AIME\)
Rollout 8
for k=2, the color sequences are RRGGG and GGGGRR. Wait, RR
Rollout 8
GGGGR\n\nk = 2: RRGGG, GGGGRR\n\nWait, wait
Rollout 9
the area common to triangle \(GEM\) and square \(AIME\) is 80 square units
Rollout 1
those y such that ||y - x|| = ||y - x'|| = d / sqrt(2
Rollout 5
) \implies 3b + 75 = 2b + 150 \
Rollout 8
either reds first then greens or greens first then reds. For each possible split of red and green
Rollout 8
2.\n\nNumber of card sequences for RRRGG: P(5,3)*P(5
Rollout 1
(2}, so that the distance between any two vectors would be sqrt( (d / sqrt(2
Rollout 8
k=2, the color sequences are RRGGG and GGGGRR. Wait, RRGG
Rollout 8
2: RRGGG, GGGGRR\n\nWait, wait, maybe I need to re
Rollout 8
k=2:\n\nEach color sequence (RRGGG and GGGGR R) corresponds to P(
Rollout 2
[\n x \equiv 5 \mod 8 \quad \text{and} \quad x
Rollout 1
\{ \frac{\sqrt{2}}{d}(x - y) : x \in S
Rollout 8
for k=2:\n\nEach color sequence (RRGGG and GGGGR R) corresponds to P
Rollout 8
= 3: RRR GG, GG RRR\n\nk = 4: RRRR G
Top 16 Negative Activations
Rollout 3
of radon gas. Hmm, first, I remember there's a formula for the root mean square speed
Rollout 3
22 g/mol, right?\n\nSo, first, M = 222 g/mol =
Rollout 6
here.\n\nHence, you just reference this result. However, since the problem asks to show it,
Rollout 3
of 222 g/mol. Therefore, the molar mass M is 222 g
Rollout 4
figures. The question might not require rounding, but to check if my initial error propagations are correct.\n\n
Rollout 3
222 g/mol, right?\n\nSo, first, M = 222 g/mol =
Rollout 5
/100. \n\nFirst, let me recall some trapezoid properties. In a tr
Rollout 5
think\nOkay, let's see. I need to solve this problem about a trapezoid with
Rollout 3
6*0.222) )\n\nFirst compute numerator:\n\n8*8.314 =
Rollout 6
that's probably the answer here.\n\nHence, you just reference this result. However, since the problem
Rollout 3
root mean square speed. But maybe in this context, they just want the root mean square speed? The
Rollout 3
isotopes like Rn-220 and Rn-219. But the most common
Rollout 3
different from the root mean square speed. But maybe in this context, they just want the root mean square
Rollout 7
) becomes 59.\n\nPart (a) is asking for all possible values of N given these conditions
Rollout 4
2 V. Hmm, alright, let's start by recalling how to calculate the equivalent capacitance of capac
Rollout 8
) but considering order.\n\nWait, but when they compute the probability, is the sample space the number of

UP_PROJ

Top 16 Positive Activations
Rollout 2
for k>=3, hence (10^k -1)≡-1 mod125
Rollout 4
, since energy is (1/2)*C_eq*V2, maximum energy would be when C
Rollout 2
25 for k>=3, hence (10^k -1)≡-1 mod1
Rollout 0
, since a is a number composed by assigning primes to a (possibly a = product of primes raised to
Rollout 6
because we can ignore the angel string (set α_n to some fixed string, which can be computed trivial
Rollout 2
5 for k>=3, hence (10^k -1)≡-1 mod12
Rollout 2
0 mod 8, which would make the entire product congruent to 0 mod 8.\n\nCompute
Rollout 6
n=2 and n=3 would give p(n) = 4 and 9, different.
Rollout 7
=8.\n\nAfter increase promoted total:79*(16 +x)=79*(16
Rollout 9
8.\n\nAt x=10: GE: y=(-1/5)(10)+2=
Rollout 9
10.\n\nAt x=0: GE: y=2, GM: y=8.\n\nAt
Rollout 0
144,5)=1, and ab=144*5=720 indeed
Rollout 7
total score of original repeaters is 56*(N/3).\n\nLet’s denote T = total
Rollout 7
their total score:\n\ntotal promoted score =79*(P +x).\n\nBut original promoted now scores higher
Rollout 2
process again.\n\nFirst modulus8:\n\nProduct=(10^1 -1)(10^2 -
Rollout 7
71N as before. Total score equals promoted total plus non-promoted total:79*(P
Top 16 Negative Activations
Rollout 9
, EM is the top side of the square. Wait, hold on. \n\nWait, in the square
Rollout 7
5, that's original score >=60. Wait, but originally, the pass mark was 6
Rollout 9
are AI, IM, ME, and EA. Wait, maybe that's a better way. Since each
Rollout 4
8%. Wait, let me check the exact value.\n\nFirst, compute (ΔC_eq / C_eq
Rollout 0
is 2, which is even. Wait a second. Then primes 11, 13
Rollout 2
to 999. Wait, no. Wait, the first term is k=1 (1
Rollout 8
* 8 * 7 * 6. Alternatively, that's 30240.
Rollout 4
which rounds to 0.88%. Wait, let me check the exact value.\n\nFirst, compute
Rollout 8
31/126.\n\nWait, wait: 7440 ��240
Rollout 8
* 8 * 7 * 6. Wait, because each card is drawn without replacement and the
Rollout 8
anywhere from 0 to 5 reds. Wait, but in the problem statement, examples include R
Rollout 8
wait, maybe I need to reorganize this.\n\nWait, the split is k reds and (
Rollout 9
EA. Wait, maybe that's a better way. Since each side is 10 units. So
Rollout 0
Wait, so maybe I need to rethink that.\n\nWait, actually 20! has prime factors with
Rollout 4
[0.003]^2. But wait:\n\nFirst term inside the formula is [ (C
Rollout 7
54)N/3 ?\n\nWait, wait:\n\nWait 18N is equal to 5

DOWN_PROJ

Top 16 Positive Activations
Rollout 9
h=10 would be (base * height)/2 = (10 * 10)/
Rollout 4
ence, [0.003]^2 = 9e-6\n\nSecond term: [
Rollout 5
that creates a certain area ratio. Let me also recall that the area of a trapezoid is
Rollout 4
2000*3000 = 6,000,000
Rollout 1
to use the parallelogram law. Let me recall that in a Hilbert space, given two vectors
Rollout 9
which is 80.\n\nTherefore, need to find h such that the area of overlap is 8
Rollout 8
's see: 240×30 =7200, 7440
Rollout 1
in S, we have certain equations.\n\nBut how do you define a centroid in infinite dimensions? The usual
Rollout 3
Let's see, 168^2 = 28,224. 1
Rollout 4
5025 * 1200 = 6,030,000
Rollout 9
50, as before. So how can we have overlapping area of 80? Since the
Rollout 1
of points at distance d / sqrt(2) from each xi would need to be equidistant from
Rollout 1
infinitely many spheres, but I might need some condition to ensure that.\n\nAlternatively, using the hypothesis that the
Rollout 1
/ sqrt(2) * sqrt(2) = d.\n\nTherefore, yeah, the setup is kind
Rollout 1
y as the "centroid" or a point such that for all x in S, we have certain equations
Rollout 7
led to impossibility. Thus, perhaps the answer is different.\n\nAlternatively, maybe N=12.
Top 16 Negative Activations
Rollout 8
adjacent). But the way the problem is phrased: "all the red cards laid out are adjacent
Rollout 5
= 0\n\nSimplify:\n\ny^2 - 150y - 125
Rollout 0
, which is odd. Actually, wait nvm, the exponents for 2,3,5
Rollout 9
from both sides:\n\n0 = 100 - 20y\n\nSo 20y
Rollout 0
9 <80: yes.\n\n4. Assign 5 to a, others to b: a
Rollout 0
16 <45: yes)\n\n3. Assign 3 to a =>a=9, b
Rollout 3
temperature, this seems ambiguous. However, in many standard problems, they might take 273 K
Rollout 7
2R\n\nSo in the original scenario, the number of promoted participants is twice the number of repeaters
Rollout 2
we assured that this is the case?\n\nWait, but this conflicts with our result for modulo8. If
Rollout 8
muting them) * C(5,5−r) *(5−r)! (choosing
Rollout 2
-120=5; 125≡5 mod8.\n\nSimilarly 109
Rollout 8
maybe I need to reorganize this.\n\nWait, the split is k reds and (5 -
Rollout 8
, for k = 1, 2 colors sequences: R GGGG, GGGG R
Rollout 3
this seems ambiguous. However, in many standard problems, they might take 273 K as ST
Rollout 0
5 <144: yes.\n\n5. Assign 2 and 3 to a: a=
Rollout 8
you first choose and arrange r red cards, then choose and arrange 5−r green cards. These

Layer 31

GATE_PROJ

Top 16 Positive Activations
Rollout 2
09. But wait, but I feel like in this case, the product of all these numbers,
Rollout 8
's 30240. But let me verify that. Yes, 10 choices for
Rollout 8
: 1200\n\nBut wait, the total for each k is 2 * [P
Rollout 3
Wait, I need to confirm. But let's see, the question is asking for the "mean molecular
Rollout 3
a standard temperature to assume. Let me check online, but since I can't do that, I need
Rollout 9
0 and x=10). Wait, but in trapezoid area formula:\n\nAverage of the
Rollout 0
.\n\nBut wait, can it be? Let me check with a smaller factorial.\n\nSuppose instead of
Rollout 3
which one is the actual mean speed. Let me think.\n\nMean speed, or average speed, is indeed
Rollout 8
and green cards to the green positions. Let me see.\n\nFor a specific color sequence, say RRGG
Rollout 0
20! has exponents:\n\nLet me do prime factorization of 20!.\n\nTo compute
Rollout 3
radon gas at a standard temperature. Let me think. If I remember, for example, for gases
Rollout 8
of permutations (10P5).\n\nLet me try this approach.\n\nFirst, how many color sequences are
Rollout 2
09 mod1000. But let me check this process again.\n\nFirst modulus8:\n\nProduct
Rollout 0
divide the total by 2.\n\nWait, but the problem says "how many rational numbers between 0
Rollout 5
181. \n\nWait, but let me verify once again. The midline divides the tr
Rollout 9
use the formula for the area between two lines. The area between the two lines from x=0 to
Top 16 Negative Activations
Rollout 6
{P}_{angel}$ if there exists a polynomial $p : \mathbb{N}
Rollout 6
}$ with a constraint that there should exists a polynomial time algorithm that can **compute** the angel string
Rollout 6
\{\alpha_n\} \) and a polynomial-time machine \( M \). By the Meyer's
Rollout 6
said to be **sparse** if there exists a polynomial $p : \mathbb{N}
Rollout 5
the midpoints of the legs, has a length equal to the average of the two bases, \( b
Rollout 6
/poly is equivalent to languages decidable by a polynomial-time Turing machine with a polynomial-length advice string,
Rollout 6
N}$ . In other words, there is a poly-time algorithm $A$ such that
Rollout 9
and square \(AIME\) is 80 square units. We need to find the length of the
Rollout 6
}^{p(n)}$ , and a deterministic polynomial time Turing Machine $M$ such that
Rollout 6
decidable by a polynomial-time Turing machine with a polynomial-length advice string, which can be arbitrary. It
Rollout 6
languages decidable by a poly-time machine with a poly-length advice string that can be generated in poly-time
Rollout 6
which is still polynomial in \( m \). The polynomial-time Turing machine \( M \) with oracle \(
Rollout 6
P_angel is. P_angel allows a poly-length advice string α_n for each input length n
Rollout 4
0000}{5000} = 1200 \, \text{
Rollout 6
$ in $S$ is bounded by $p(n)$ . In other words,
Rollout 7
mark is fixed at 65. The original average scores are 66 for all participants,

UP_PROJ

Top 16 Positive Activations
Rollout 3
Rn, which is approx. 1/8 the speed (mass is ~8 times higher,
Rollout 7
(R - x).\n\nTherefore, A*x + B*(R - x) = 56R
Rollout 9
80, so maybe the triangle is partially outside and partially inside. If point G is outside the square
Rollout 9
(1/5 x +8 +1/5 x -2 ) ] dx\n\n= ��
Rollout 7
P=2N => P = (2/3)N.\n\nSo N must be divisible by
Rollout 7
their averages increased by 5, since they combine with x students whose original scores were between 60
Rollout 7
terms:\n\n50N +59N/3 + (75x -59x)=
Rollout 8
=0: 2!; total 2+2=4.\n\nCase2: for r=
Rollout 3
for Rn, which is approx. 1/8 the speed (mass is ~8 times higher
Rollout 9
is too small. Hence, h must be less than 10, making G located inside the square
Rollout 5
/100 is 181.25, so the greatest integer not exceeding that is
Rollout 9
[ (1/5 x +8 +1/5 x -2 ) ] dx\n\n= �
Rollout 1
)1.5d2 = -0.5d2.\n\nThis is not zero. Hence,
Rollout 9
[0 to10] [ (2/5 x +6 ) ] dx\n\n= [ (
Rollout 3
222 vs. 28≈8*28=224. So 2
Rollout 9
=10: GE: y=(-1/5)(10)+2= -2 +2
Top 16 Negative Activations
Rollout 9
25.\n\nPerfect.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(A
Rollout 5
the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n<|im_start|>answer\nLet the
Rollout 3
69 m/s.\n\n**Final Answer**\n\boxed{170} m/s\n\nWait no
Rollout 0
}=128.\n\n**Final Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven a
Rollout 2
109.\n\nTherefore, boxed answer is \boxed{109}\n\n**Final Answer**\n\
Rollout 2
{109}\n\n**Final Answer**\n\boxed{109}\n\n<|im_start|>answer\nTo find
Rollout 8
permutations, so 10P5. Yeah.\n\nBut perhaps another perspective: if she's selecting
Rollout 8
wait, maybe it's better to think:\n\nWait, arranging r reds and 5−r greens
Rollout 8
s and 4 to 1 greens).\n\nWait, so for each possible k (number of reds
Rollout 5
181 is the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n
Rollout 5
5 is 181.\n\n\[\n\boxed{181}\n\]<|im_end|>
Rollout 2
divided by 1000 is \(\boxed{109}\).<|im_end|>
Rollout 8
157.\n\n**Final Answer**\n\boxed{157}\n\n<|im_start|>answer\nKathy
Rollout 5
* sqrt(725). Is that right?\n\nWait, let me check the calculations again step by
Rollout 9
, so that would be the vertical side. Wait, no, in that labeling, if AIME is
Rollout 9
in \(\triangle GEM\) is \(\boxed{25}\).<|im_end|>

DOWN_PROJ

Top 16 Positive Activations
Rollout 7
32x=8N/3 => x=(8N)/(3*32)=N/
Rollout 5
(x + 175)H [1 - 125/(75 + x)]
Rollout 5
5H\n\nBut from the first equation: h = (125H)/(75 + x
Rollout 9
20=500/h\n\nThen h=500/20=25.\n\n
Rollout 9
). Since 50/h and 10 -50/h.\n\nNow, but if 1
Rollout 7
16x=4N/3 => x= (4N)/(3*16)=N
Rollout 7
2/12=14/3 +3.5=14/3 +7/
Rollout 0
a^2 < 20! or a < sqrt(20!). Therefore, the number of
Rollout 1
how can we construct such a y?\n\nAlternatively, perhaps by considering properties of such sets S. If S
Rollout 6
their bin(n) representations have different length. But maybe if we fixed the encoding to use the same number
Rollout 6
on the growth of p(n). For example, if p(n) is monotonic and grows polynomially
Rollout 0
number of such a is half the total number of divisors, adjusted for square factors. However, we
Rollout 1
Denote t = 2y. Then y = t/2.\n\nFrom the sphere condition, ||
Rollout 7
N --> 3P=2N => P= (2/3)N.\n\nYes that's
Rollout 3
Rn, which is approx. 1/8 the speed (mass is ~8 times higher,
Rollout 9
500/h=20, so h=25. If h=25, then
Top 16 Negative Activations
Rollout 3
in kinetic theory, they are different. Let me make sure.\n\nAccording to Maxwell-Boltzmann distribution
Rollout 4
approximately 0.36%, consistent with the previous result.\n\nTherefore, combining this with the 2
Rollout 5
= 181.25, so the greatest integer not exceeding this is 181
Rollout 2
)≡109 mod125. Correct.\n\nThen solve for x≡5 mod8 and
Rollout 5
now, but maybe I had a miscalculation in the algebra step. Let me verify quickly.
Rollout 2
the same final result, so now, why did my modulus8 and modulus125 approach via CRT
Rollout 1
/ 2 = 0.\n\nCompare to the above line:\n\nFrom the sum of norms, we have
Rollout 2
to modulo8 analysis? Let's be precise.\n\nbreak down the terms in the product when mod12
Rollout 5
the calculations again step by step. Probably, I made an algebraic mistake.\n\nStarting from the two equations
Rollout 2
inconsistencies.\n\nHence, getting109 in both cases. So The answer is109.
Rollout 7
=209N/3.\n\nSimilarly, the x terms:\n\n(75 -59)x
Rollout 7
56*4=224. Correct.\n\nAfter increase, promoted total score: original promoted
Rollout 2
^4 -1; multiplying by that, so previous result109 * (10^4
Rollout 3
the question refers to the "mean" speed, it's probably the average speed formula. But maybe I
Rollout 7
score increase, the average of the promoted participants (which includes original promoted and the x repeaters) is
Rollout 0
But 2^{3-1}= 4 correct. Hence, Therefore, the formula is 2

Layer 32

GATE_PROJ

Top 16 Positive Activations
Rollout 6
\) unless \( \textbf{NP} = \textbf{P} \).\n\n---\n\n**
Rollout 2
91 mod125: 125*7=875. 89
Rollout 4
{C_1 \cdot C_2}{C_1 + C_2}\n\]\n\nGiven
Rollout 2
91. 891 mod125: 125*7=87
Rollout 2
9's}}$ is divided by $1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright
Rollout 4
eq}} \) and its uncertainty. The equivalent capacitance for capacitors in series is given by:\n\n
Rollout 2
91: 891mod125. 125*7=87
Rollout 2
=999≡-1 mod125.\n\nIn fact, as noticed before, 1
Rollout 4
10 pF and C2 = 3000 ± 15 pF, and
Rollout 6
\), unless \( \textbf{NP} = \textbf{P} \), this equality
Rollout 7
0. 36 is less than 40, so all three are acceptable.\n\nWait, so
Rollout 8
= 10! / (10-5)! = 10 * 9 *
Rollout 2
the problem modulo 8 and modulo 125 separately, since 1000 is
Rollout 2
891. 891 mod125: 125*7=8
Rollout 8
of distinct color sequences is C(10,5) but considering order.\n\nWait, but when they
Rollout 2
91. 891 mod125=16, then 16*(-1
Top 16 Negative Activations
Rollout 0
1.\n\nSo primes 2: 18, 3: 8, 5:4
Rollout 4
significant figures, so 0.88%, or maybe the answer is exact at 0.8
Rollout 3
is 168.5 m/s, and rounded to three sig figs is 16
Rollout 0
corresponding to \boxed{128}? Or wait, wait prime factors of 20!
Rollout 0
So exponents are: 18 (2), 8 (3), 4 (5),
Rollout 4
Then total relative variance in C_eq is 9e-6 + 4e-6 =
Rollout 3
atomic number is 86, so atomic weight is approximately 222 (since the most stable
Rollout 9
also, since EM is vertical, G is somewhere to the left or right of EM. Since the triangle
Rollout 1
(d, squared + ||y|| squared ) / 2 = (d2 + d2 /
Rollout 2
97} mod125\n\nBecause from k=3 to k=999, that
Rollout 0
\(a \neq b\). Hence, total coprime pairs are \(2^{8}\
Rollout 3
168.5≈ 168 or 169. In technical terms,
Rollout 4
,150 /5025 = approximately 6. So total is 120
Rollout 2
000. As997 terms fromk=3 tok=999. 9
Rollout 8
card is drawn without replacement and the order matters. Alternatively, the number is 10P5 =
Rollout 0
} ordered coprime pairs (a,b), and because n! is not a square (unless for

UP_PROJ

Top 16 Positive Activations
Rollout 3
*R*T has units (kg·m2)/(s2·mol). Then, dividing by M in
Rollout 1
non-empty?\n\nThere's a theorem in the theory of Hilbert spaces called the Kakutani's theorem
Rollout 6
= O(log n), how many strings of length m are in S_L? Since m is log n
Rollout 1
if we can identify y as the center of such a system. The problem is the existence of y.\n\n
Rollout 6
p(n), that corresponds to one string of length m per n equal to p^{-1}(m).
Rollout 6
_i has at most p(n) strings of length n. Since there are k sparse sets, maybe we
Rollout 1
. If S is a set of points each lying on the sphere of radius d/sqrt(2)
Rollout 1
\{ \frac{\sqrt{2}}{d}(x - y) : x \in S
Rollout 6
overlapping). Hence, the number of strings of length m is O(1), which is polynomial (constant
Rollout 6
Answer:** \n\( \textbf{P} = \textbf{P}_{\text{bad
Rollout 6
a polynomial value, the number of strings of length m is at most 1, which is sparse.
Rollout 8
30240 ��240= 126. So 31
Rollout 1
each pair is separated by d, perhaps S must lie on such a quadratic manifold. But unless I can
Rollout 6
. Therefore, \( \textbf{P} = \textbf{P}_{\text{bad
Rollout 8
31, 30240÷240=126. As above.
Rollout 6
oracle, and P_angel is basically P/poly, so the answer is invoking this theorem. Thus
Top 16 Negative Activations
Rollout 0
,5,7 have exponent 1, and exponents for 2,3,5,7
Rollout 2
1≡16 mod125\n\nSo now we have 16 * (-1)^{
Rollout 5
75) = 2/3. Cross-multiplying: 3(b + 25
Rollout 0
and \(b = 20!\), or vice versa; however, \(a\) and \(b
Rollout 0
\(b\). So for each prime factor, we assign it to either \(a\) or \(b
Rollout 0
13,17,19 are 1 (each odd?), and the others are even.
Rollout 2
$ is divided by $1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I
Rollout 0
primes. However, since for each such assignment, either a < b or a > b. Since the
Rollout 5
5) = 2/3. Cross-multiplying: 3(b + 25)
Rollout 4
could use logarithmic differentiation. Taking the natural logarithm of C_eq: ln C_eq = ln C
Rollout 0
each of the prime powers in the factorization of 20! must go entirely to one of the
Rollout 4
2) - ln(C1 + C2)\n\nDifferentiate both sides:\n\ndC_eq / C_eq
Rollout 2
by $1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem
Rollout 0
20! has prime factors with exponents. For example, 2^18, 3
Rollout 9
))^2 + (10 - y)^2] = sqrt[h^2 + (10 -
Rollout 2
{997} = -1 mod8, because 997 is odd.\n\nTherefore,

DOWN_PROJ

Top 16 Positive Activations
Rollout 9
celes triangle with base EM: if it's vertical, then the base is vertical. But triangles are
Rollout 9
triangle GEM with base EM. So EM is one of the sides of the square, the bottom side
Rollout 9
triangle. The triangle is isosceles with base EM, so the legs would be GE and GM
Rollout 9
apex of the triangle opposite the base EM, which is the vertical side of the square. If the altitude
Rollout 9
GEM is an isosceles triangle with base EM, so EM is from (10,
Rollout 9
. Since the triangle is isosceles with base EM, the apex is G. And the altitude
Rollout 9
the triangle opposite the base EM, which is the vertical side of the square. If the altitude is,
Rollout 9
base EM: if it's vertical, then the base is vertical. But triangles are usually thought of as
Rollout 9
: if it's vertical, then the base is vertical. But triangles are usually thought of as having a
Rollout 9
). But the triangle being isosceles with base EM. Since triangle GEM is isosce
Rollout 9
with base EM. So EM is one of the sides of the square, the bottom side from E to
Rollout 9
have an isosceles triangle GEM with base EM. So EM is one of the sides of
Rollout 9
the left of EM. If triangle GEM has base EM (at x=10), and is
Rollout 8
ones) are adjacent and all green cards (regardless of which ones) are adjacent. Therefore, the
Rollout 9
wait, the problem says the base of the triangle is EM, which in this case is the vertical side
Rollout 9
Wait, no. Wait, EM is the top side of the square. Wait, hold on. \n\n
Top 16 Negative Activations
Rollout 1
l is non-empty if and only if r1 + r2 >= l and |r1 - r
Rollout 1
that even for three points, you can find such a y. But in higher dimensions, there's more
Rollout 1
2, the intersection is non-empty if r1 + r2 >= d. In our case, r
Rollout 6
sets with few strings per length. The idea might be to encode the advice strings α_n into S_L
Rollout 2
8. Since 5 and8 are coprime, m must be ��0 mod8.
Rollout 6
But since α_n is length p(n), perhaps use p(n) strings, each encoding a bit of
Rollout 5
= \frac{2}{3}\n\]\nSolving this, we find:\n\[\n3(b
Rollout 1
d2 / 2. Which can be considered as equations in H for y.\n\nNow, the set
Rollout 6
is standard.\n\n<|im_start|>answer\n**Problem 1:**\n\nTo construct the sparse set \( S \),
Rollout 9
, y) to E (10,0).\n\nDistance GE = sqrt[(10 - (1
Rollout 7
4) = 2N/3\n\nDivide both sides by N (assuming N ≠0,
Rollout 7
)(A -79)=0\n\nDivide by N (N≠0):\n\n-2 + (
Rollout 5
on the average of the bases times the height, to get equal areas, the ratio of the heights squared
Rollout 2
0 mod8. So 5m is divisible by8. Since 5 and8 are copr
Rollout 7
294. Which is 294/3=98 per student, but it's
Rollout 7
must be integer, so N must be a multiple of 12. Just like part (a).

Layer 33

GATE_PROJ

Top 16 Positive Activations
Rollout 3
8,436). Let's see, 168^2 = 28,
Rollout 3
8,408). Let's see.\n\n168^2 = (170 -
Rollout 3
Alternatively, precise sqrt(26050)=161.4, as 16
Rollout 4
0*2985. Let's compute that.\n\n1990 * 298
Rollout 3
sqrt(26,049) �� 161.4 m/s. So
Rollout 3
6). Let's see, 168^2 = 28,224.
Rollout 3
33,476.2) �� 183 m/s. So that's
Rollout 3
sqrt(28,433) �� 168.6 m/s.\n\nSo
Rollout 3
50)=161.4, as 161^2=25,9
Rollout 8
0. Let's see: 240×30 =7200, 7
Rollout 3
.\n\nsqrt(28,436). Let's see, 168^2 =
Rollout 3
8,408.12) �� 168.5 m/s\n\nSo
Rollout 3
3). Let's compute with square root method:\n\n168.0^2=28,
Rollout 3
2 = 28,224. 169^2 = 28,
Rollout 3
28,436). Let's see, 168^2 = 28
Rollout 3
=28,224.0\n\n168.5^2=168
Top 16 Negative Activations
Rollout 4
C1 ΔC2 ) / (C2 (C1 + C2)) ] = (2
Rollout 7
Therefore,\n\nx(A -54)=2N/3.\n\nBut A is the original average of the
Rollout 4
C2 ΔC1 ) / (C1 (C1 + C2)) ] = (3
Rollout 3
8*R*T has units (kg·m2)/(s2·mol). Then, dividing by M
Rollout 3
= sqrt((8 * R * T) / (π * M))\n\nR = 8.3
Rollout 3
in denominator; thus overall (kg·m2)/(s2·mol) divided by (kg/mol
Rollout 1
sqrt(2). Then scaling by sqrt(2)/d would make them orthonormal.\n\nSo the
Rollout 3
overall (kg·m2)/(s2·mol) divided by (kg/mol) is (m
Rollout 5
k = (a + b)h / 2. But we also need to relate k and h
Rollout 4
1 ]^2 + [ (C1 / (C2 (C1 + C2))) Δ
Rollout 4
is [ (C2 ΔC1 ) / (C1 (C1 + C2)) ]
Rollout 4
expression:\n\nFirst term: [ (C2 / (C1 (C1 + C2))) Δ
Rollout 7
,\n\nx(A -54)=2N/3.\n\nBut A is the original average of the x
Rollout 4
/ (C1 (C1 + C2)) ] = (3000 * 1
Rollout 4
C1 ΔC2 ) / (C2 (C1 + C2)) ]^2 \n\n
Rollout 1
the main idea is to find a point y such that all x in S are at distance d/sqrt

UP_PROJ

Top 16 Positive Activations
Rollout 4
0036058 as a decimal, so yes, 0.3605
Rollout 3
)/337≈0.047. Thus, sqrt≈168.5 +
Rollout 7
9/6≈8.1667\n\nThus:\n\nA=12*(49/
Rollout 9
10 -20 = -10, left outside the square. So if h>1
Rollout 3
/337≈0.047.\n\nTherefore, sqrt≈168.5 +
Rollout 3
9,953.6 minus 66.512*2=133
Rollout 7
and x must be <= R = N/3. Also, the original repeaters include x students with
Rollout 3
3.6 minus 66.512*2=133.024
Rollout 9
10 - h)) = (-5)/h.\n\nThe equation is y -0 = (-5)/
Rollout 4
60583333... So approximately 0.3606% as relative
Rollout 4
3.1623e-3, so 2.774 * 3
Rollout 3
15.87 /337≈0.047.\n\nTherefore, sqrt≈
Rollout 7
/3. Since x= N/12, this implies N must be divisible by 12
Rollout 7
must be between 60 and 64. Therefore, A can be from 60 to
Rollout 3
87 /337≈0.047.\n\nTherefore, sqrt≈168.
Rollout 7
09N)/3=4N/3.\n\nTherefore:\n\n16x=4N/3
Top 16 Negative Activations
Rollout 5
00 \). The midline of the trapezoid, which connects the midpoints of the
Rollout 5
\) of the segment that divides the trapezoid into two regions of equal area, we use
Rollout 6
sets \( S_1, S_2, \ldots, S_k \) by tagging each
Rollout 0
) and \(\gcd(a, b) = 1\), each prime factor of 20
Rollout 7
59 for non-promoted participants. \n\n### Part (a)\n\n1. **Original equations:
Rollout 7
= 2R \) and \( N = 3R \), so \( N \) must
Rollout 6
not provide the nondeterminism inherent in \( \textbf{NP} \).\n\n**Answer:** \n
Rollout 8
\): \(2 \times (5 \times 4 \times 5 \times 4 \
Rollout 1
y) \) and \( \frac{\sqrt{2}}{d}(x' - y)
Rollout 7
<|im_start|>answer\nGiven a mathematics test with \( N \) participants (\( N < 40 \
Rollout 1
\) with radius \( \frac{d}{\sqrt{2}} \) can be non-empty
Rollout 8
10, which is \(10 \times 9 \times 8 \times 7 \
Rollout 3
olar mass of approximately \(222 \, \text{g/mol}\), which converts to \(
Rollout 9
a coordinate system with vertices \(A(0, 0)\), \(I(0, 1
Rollout 5
length \( x \) of the segment that divides the trapezoid into two regions of equal area
Rollout 7
10R\n\nDivide both sides by 5:\n\nP = 2R\n\nSo in

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
28,224.\n\n168.5^2= (168 +
Rollout 3
,224.0\n\n168.5^2=168^2 +
Rollout 5
100 = 181.25, so the greatest integer not exceeding this is
Rollout 3
is four significant digits (if we consider the trailing.5 as a justified digit). However, based on
Rollout 4
30,000 and 10*15=150. So total would
Rollout 4
64% ��0.8775%\n\nBut maybe doing precise sqrt(0.0
Rollout 4
0,250. 5,940,150 -5,92
Rollout 7
impossible. Hence under any circumstances,given these constraintsthe answer to part(b) is that there is
Rollout 4
0.877 is 0.88 when rounded to two decimal places. \n\nAlternatively,
Rollout 3
2.6 - 24.942*2=49.884,
Rollout 4
74%, which rounds to 0.88%. Wait, let me check the exact value.\n\n
Rollout 4
\n\nAnd (0.003606)^2 �� (0.003
Rollout 3
02.4\n\n66.512 * 90 = 5,98
Rollout 3
are 8 (exact), R=8.314 (4), T=298
Rollout 3
90 +3)=66.512*200=13,30
Rollout 3
3.6 - 66.512*7=465.584
Top 16 Negative Activations
Rollout 3
is a gas at standard temperature and pressure, with boiling point at -61.7 °C,
Rollout 5
bases that divides the trapezoid into two regions of equal area, its length x should be such
Rollout 5
the legs divides the trapezoid into two regions with areas in the ratio 2:3.
Rollout 5
bases that divides the trapezoid into two regions of equal area, the length x can be found
Rollout 8
of red and one block of green (which could be in either order), or all reds or all
Rollout 9
be a large triangle with base EM (the right edge of the square) and apex G outside the square
Rollout 9
10 -100/h)*x ] from 0 to10.\n\nEvaluating at
Rollout 5
midline divides the trapezoid into two regions with areas in the ratio 2:3.
Rollout 9
Wait, no. Wait, EM is the top side of the square. Wait, hold on. \n\n
Rollout 9
with base EM. So EM is one of the sides of the square, the bottom side from E to
Rollout 9
x^2/2) +6x ] from0 to10\n\n= [ (1/
Rollout 5
divides the trapezoid into two regions with areas in the ratio 2:3. Then,
Rollout 9
apezoid with bases at x=0 (length 8-2=6) and x=
Rollout 5
segment that divides the trapezoid into two regions of equal area. \n\nFor a line segment parallel
Rollout 7
oted (repeaters) becomes 59.\n\nPart (a) is asking for all possible values of
Rollout 5
divides the trapezoid into two regions with areas in the ratio 2:3. \n\nWait

Layer 34

GATE_PROJ

Top 16 Positive Activations
Rollout 9
is 100 -500/h=80. Then 500/h=
Rollout 9
(10,0):\n\nUsing point-slope form: y -0 = (-5/h)(x
Rollout 7
P = 56N +15P=66N\n\nSo 15P=
Rollout 3
standard temperature to assume. Let me check online, but since I can't do that, I need to
Rollout 5
up the equation: (b + 25)/(b + 75) = 2/
Rollout 5
150y -12500 = 0\n\nSolving for y using quadratic formula
Rollout 7
N/3)=59N/3\n\nCombine like terms:\n\n50N +59N
Rollout 6
if the ith bit of α_n is b. But similar to previous thoughts.\n\nBut then S_L could
Rollout 5
500 = 125y\n\nBring all terms to left side:\n\ny^2 -
Rollout 5
75 from the midline area ratio 2:3.\n\n2. Remembered that the line dividing
Rollout 1
we can see this as a functional equation. Let's denote f(x) = 2⟨x,
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $A
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics
Top 16 Negative Activations
Rollout 2
^1 -1), the second is k=2 (10^2 -1), ..., up
Rollout 2
1)=9 mod125\n\nk=2:10^2 -1=99
Rollout 2
2 -1 =1 mod8\n\nk=2: (4 -1)=3 mod8\n\n
Rollout 2
modulus125, each term from k=3 onward is≡-1, so multiplying 9
Rollout 2
:10 -1=9.\n\nk=2:100 -1=99.\n\n
Rollout 2
00 -1=99.\n\nk=3:1000 -1=99
Rollout 2
first two terms (k=1 andk=2) are9,99. The remaining
Rollout 2
=99 mod125\n\nk=3:10^3 -1=99
Rollout 2
}^{999} (10^k -1). So we're dealing with a product
Rollout 2
product from k=1 to999 are even; for example, 9 is odd (mod
Rollout 2
0^1 -1=9;\n\nk=2,10^2 -1=99
Rollout 2
} mod125\n\nBecause from k=3 to k=999, that's
Rollout 2
^2 -1=99\n\nk=3, so10^3 -1=9
Rollout 2
in the product is a number made up of 9s. Let's figure out how each term behaves
Rollout 6
textbf{NP}$ For example, all unary languages, even $UHALT$
Rollout 2
:2-1=1. For k=2:4 -1=3. For k>=

UP_PROJ

Top 16 Positive Activations
Rollout 8
original problem, seems that my answer is right.\n\nTherefore, probability is 31/126
Rollout 8
800 = 7200\n\nTherefore, total happy sequences:\n\nCase1 + Case2
Rollout 2
methods confirm that the remainder is109.\n\nTherefore, answer:109.\n\nBut let me
Rollout 3
0.222) )\n\nFirst compute numerator:\n\n8*8.314 = 6
Rollout 2
91≡16 mod125\n\nSo now we have 16 * (-1)^
Rollout 2
5=109 mod125.\n\nTherefore, P ��109 mod12
Rollout 0
19. So that's 8 primes. Therefore, the number of coprime pairs would be
Rollout 3
. So approx 161 m/s.\n\nSo depending on the temperature assumption, the answer would be
Rollout 2
) = -3 ��5 mod8.\n\nTherefore, P �� 5 mod8.\n\nAlright
Rollout 5
25y\n\nExpanding the left-hand side:\n\ny^2 - 125y +
Rollout 4
So approximately 0.8776%.\n\nThus, the percentage error is approximately 0.8
Rollout 3
figs justify 169 m/s.\n\nTherefore the valid boxed answer is \boxed{16
Rollout 3
820.576\n\nSo numerator is 19,820.57
Rollout 5
integer not exceeding that is 181. Therefore, the answer is 181. \n\n
Rollout 5
what we have. Thus, that is consistent.\n\nTherefore, x2 is 18125
Rollout 0
formula is 2^{k -1).\n\nHence, for 20! with 8 prime
Top 16 Negative Activations
Rollout 8
available.\n\nTherefore, this is similar to permutation with two colors. The number of such sequences is C(
Rollout 1
orthogonal to the set S - S, except for a constant.\n\nBut in that case, if we take
Rollout 1
with equal norms and their differences have equal norms, this often relates to them being vertices of a simplex or
Rollout 1
of them.\n\nAlternatively, for the system to have a solution, then for any x, x' in
Rollout 1
them as orthogonal vectors. The proof existence uses the separability and completeness.\n\nBut maybe an innovative approach is
Rollout 1
2, for all x in S, has a common solution y in H. Alternatively, the intersection
Rollout 1
- y|| = d / sqrt(2) } is not convex. Oh, right. The sphere
Rollout 1
and their differences have equal norms, this often relates to them being vertices of a simplex or something similar.
Rollout 1
intersection is non-empty. So, since each sphere is closed and convex, and in a Hilbert space
Rollout 8
n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m +
Rollout 1
equal norms, this often relates to them being vertices of a simplex or something similar. For example, in
Rollout 1
example, in finite dimensions, if you have points on a sphere such that the distance between any two is
Rollout 1
to the set S - S, except for a constant.\n\nBut in that case, if we take x
Rollout 1
is a set where all the distances are equal, it's called an equidistant set. In Hil
Rollout 7
problem probably expects some answer, possibly also N=12,24,36, but perhaps
Rollout 7
possibly also N=12,24,36, but perhaps adjusting some assumptions.\n\nWait,

DOWN_PROJ

Top 16 Positive Activations
Rollout 9
10 -100/h)*x ] from 0 to10.\n\nEvaluating at
Rollout 9
x^2/2) +6x ] from0 to10\n\n= [ (1/
Rollout 9
the altitude to EM is the perpendicular distance from G to EM. Since EM is the vertical line from (
Rollout 9
, as I thought, the horizontal distance from G to EM (the line x=10), then
Rollout 9
= (5/h)x - (50/h) +10.\n\nWhich is y = (5
Rollout 9
But wait, if h is the altitude from G to EM, h can be anything. Let's think
Rollout 3
to message user to specify temperature, but as IA cannot, so need to make a call. \n\n Since
Rollout 9
, then if h is the distance from G to EM, then if G is inside the square, h
Rollout 2
1 mod125: 125*7=875. 891
Rollout 9
= (-5/h)x + (50/h) +0? Wait, let's verify:\n\nFrom
Rollout 9
Wait no, if the altitude is from G to EM, which is length h, then if h is
Rollout 9
. Wait no, if the altitude is from G to EM, which is length h, then if h
Rollout 9
0, the distance from (0,5) to line x=10 is indeed 10
Rollout 9
if EM is vertical, then the altitude from G to EM is horizontal. So since EM is vertical,
Rollout 9
as I thought, the horizontal distance from G to EM (the line x=10), then since
Rollout 9
0/25= (-1/5)x +2.\n\nGM: y=(5/25
Top 16 Negative Activations
Rollout 7
12,:\n\n-3*(2N/3)= (74 -A)*(N/1
Rollout 4
F}$ are connected in series. The voltage applied across this combination is $V=5.00
Rollout 7
14N/3 +42*(N/12)\n\nDivide both sides byN:\n\n
Rollout 3
= 28,561. So sqrt(28,436) is approximately
Rollout 7
Plug them in:\n\n-3*(2N/3) + (N/12)(A -
Rollout 7
/ (N/12)= (14/3)*(12/N)*(N)=14
Rollout 0
) + floor(20/4) + floor(20/8) + floor(2
Rollout 7
/3\n\nSimplify 54*N/3=18N.\n\n56N/3
Rollout 0
! for primes N/2 < p_0 <=N). For example, primes between N/2
Rollout 0
2: floor(20/2) + floor(20/4) + floor(2
Rollout 4
5000) = (30,000)/(10,000
Rollout 4
00*3000 = 6,000,000; 6
Rollout 5
= 125\n\nMultiply through by y to eliminate the denominator:\n\ny(y + 10
Rollout 7
N/3) = (75*2)/3 *N=150/3 *
Rollout 7
to N/3. Since x= N/12, this implies N must be divisible by
Rollout 4
Divide by 100: 4.327 / 12 = 0

Layer 35

GATE_PROJ

Top 16 Positive Activations
Rollout 0
answer is 128, corresponding to \boxed{128}? Or wait, wait
Rollout 4
2 + (2 * 0.004)^2 )\n\nCalculating 2*0.
Rollout 4
90=4975*(1200-10)=4975*
Rollout 3
the mean molecular speed of radon gas is \boxed{170} m/s if we round
Rollout 9
answer is 25. Therefore box form \boxed{25}\n\nWait, that seems correct.\n\n
Rollout 4
error in the calculation of the energy stored is \boxed{0.88\%}.\n\n<|im_start|>
Rollout 3
K (unless stated otherwise), perhaps answer around \boxed{169} m/s. But need
Rollout 0
<=N). For example, primes between N/2 and N cannot come in pairs. So hence ex
Rollout 3
m/s. Thus, present answer boxed as \boxed{169 \text{ m/s}}
Rollout 4
that.\n\n1990 * 2985= (2000 -10
Rollout 5
The midline length is (b + (b + 100))/2 = (2b
Rollout 2
109.\n\nTherefore, boxed answer is \boxed{109}\n\n**Final Answer**\n\
Rollout 3
19,944 – 132.96=19,811
Rollout 1
exists. Thus, finally, the answer is "\boxed{y \text{ exists as described}}" or
Rollout 3
}\).\n\n**Final Answer:**\n\[\n\boxed{169} \, \text{
Rollout 2
divided by 1000 is \(\boxed{109}\).<|im_end|>
Top 16 Negative Activations
Rollout 4
\n\nHmm, let's compute these step by step.\n\nFirst, compute the coefficients for ΔC_eq.\n\n
Rollout 1
||y||2 -d2/2.\n\nTherefore, each inner product of xi with y is this
Rollout 2
need to compute P mod 1000.\n\nNow, let's approach this modulo 8 and
Rollout 1
inner product of xi with y is this value.\n\nSo, assuming that all such equations hold.\n\nBut if
Rollout 8
compute this for k=1 to k=4.\n\nk=1:\n\n2 * P(5,
Rollout 1
⟩ - ⟨x, x⟩ = k.\n\nTherefore, 2⟨x, y⟩ = ⟨
Rollout 1
, we might use this to relate their norms.\n\nWait, but unless all the points in S have the
Rollout 2
1 to 999 mod 8.\n\nHmm. Let's compute this product. But before multiplying
Rollout 7
equation with x and N. Let's compute that.\n\nFirst expand:\n\n75*(2N/3
Rollout 6
), but this is still not polynomial per length. Wait.\n\nAlternatively, let's encode the entire α_n
Rollout 9
lines from x=0 to x=10.\n\nAt x=0: GE: y=2
Rollout 1
⟩ = ⟨x, x⟩ + k.\n\nThus, for all x in S, the linear functional
Rollout 0
many of those assignments result in a < b?\n\nSo for 720, with prime factors
Rollout 9
the square requires calculating the region inside both.\n\nHmm. Let's visualize that. The triangle has a vertex
Rollout 9
= (-5/h)x +50/h.\n\nTherefore, the area between is:\n\n∫[0 to
Rollout 9
projection path, etc. Wait, perhaps getting complicated.\n\nAlternatively, maybe split the overlapping area into the part

UP_PROJ

Top 16 Positive Activations
Rollout 7
?\n\nWait, wait:\n\nWait 18N is equal to 54N/3. So
Rollout 1
2 = 3d2/4 - which is problematic.\n\nHowever, at the same time,
Rollout 6
are in $\textbf{P}_{angel}$ because the \textit{angel string}
Rollout 1
2 = 3d2/4 - which is problematic.\n\nHowever, at the same time, the
Rollout 7
1R = 66N +5N =71N.\n\nBut separately, we can say
Rollout 7
. Which is 56*4=224. Correct.\n\nAfter increase, promoted total score
Rollout 7
, wait:\n\nWait 18N is equal to 54N/3. So 5
Rollout 3
.\n\nR is the gas constant, which is 8.314 J/(mol·K).\n\n
Rollout 1
= (||t||2)/4 = k + (d2)/2, so that k =
Rollout 3
Rn, which is approx. 1/8 the speed (mass is ~8 times higher,
Rollout 7
62=224. Which is 56*4=224. Correct.\n\n
Rollout 3
to sqrt(molar mass). So sqrt(8)=2.828. 475
Rollout 1
2r = 2d / sqrt(2) = d sqrt(2) > d, so
Rollout 7
3\n\nSimplify 54*N/3=18N.\n\n56N/3 =
Rollout 1
x, while ⟨x, x⟩ + k is quadratic. Unless all x in S lie in a
Rollout 3
However, in calculator, sqrt(28408.12)= 168.
Top 16 Negative Activations
Rollout 5
+ 175)(125H)/(75 + x) = 125
Rollout 5
125(x + 175)]/(x + 75) = 12
Rollout 9
. However, an isosceles triangle with base EM: if it's vertical, then the base
Rollout 0
a}{b}\) where \(0 < a < b\) and \(a\) and \(b\)
Rollout 9
triangle. The triangle is isosceles with base EM, so the legs would be GE and GM
Rollout 0
\(a\) and \(b\) are coprime positive integers. The product \(a \times b
Rollout 5
125(x + 175)/(x + 75)\n\nLet me write the
Rollout 0
= 20!\), \(0 < a < b\), and \(\gcd(a, b
Rollout 0
b = 20!\) and \(a < b\).\n\nNow, how do I count such
Rollout 9
, the area of overlap between triangle GEM and square AIME is 80, which is
Rollout 0
20!\), \(0 < a < b\), and \(\gcd(a, b) =
Rollout 9
and M are on the square, and G is inside, so triangle is within. But then how come
Rollout 0
20! where each pair consists of two coprime numbers.\n\nBut since the fraction is between
Rollout 0
20!. So that equivalent to saying find all reduced fractions a/b where 0 < a/b <
Rollout 0
such coprime pair where \(a < b\), we have a distinct rational number, and each
Rollout 0
times b = 20!\), \(0 < a < b\), and \(\gcd(a

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
=8.31 instead of 8.314. Let me check. Suppose R=
Rollout 7
score must be 98/1=98, but >=60 and <65.
Rollout 7
8. But the original scores can’t be98, since they need to be between60-
Rollout 2
9*1000=9999-9000=999.
Rollout 9
from (0,0) to (0,10)). Since G is to the left of the
Rollout 2
mod1000= 9999-9*1000=99
Rollout 2
99 \times \cdots \times \underbrace{99\cdots9}_{\
Rollout 4
2.774 * 3.1623e-3 �� 8
Rollout 2
*37.\n\nFourth term:9999=999*10 +9,
Rollout 3
0 +70 +3)*66.512 =\n\n200*66.
Rollout 3
K=273 with R=8.314:\n\nWe did 161 m
Rollout 2
+9, wait, perhaps 9999=9*1111, 1
Rollout 3
. Let me check. Suppose R=8.31:\n\n8*8.31=6
Rollout 3
2*298 ��24.942*300=7,4
Rollout 3
\n\nSum:13,302.4 +5,986.08=
Rollout 7
. While student’s original score must be 98/1=98, but >=60
Top 16 Negative Activations
Rollout 7
3P=2N => P= (2/3)N.\n\nYes that's correct. So
Rollout 8
5−r} or G^{5−r} R^r. So two options.\n\nIn terms
Rollout 3
25,921;162^2=26,244. So
Rollout 5
, we need to find x, which is the length of the segment that divides the trapezoid
Rollout 7
.\n\nSimilarly, the remaining repeaters are (N/3 -x) = (N/3 -
Rollout 8
of greens, either reds first then greens or greens first then reds. For each possible split of
Rollout 7
3P=2N => P = (2/3)N.\n\nSo N must be divisible by
Rollout 8
red block then green block, or first the green block then red block. Therefore, for k =
Rollout 8
2, same for all green. For the mixed color sequences (either RG or GR):\n\nThey require that
Rollout 5
we need to find x, the length of the segment that divides the trapezoid into two regions
Rollout 5
of equal area. \n\nFor a line segment parallel to the bases that divides the trapezoid into
Rollout 0
is between 0 and 1, \(a < b\). So for each such coprime
Rollout 7
with P + R:\n\n71P + 56R = 66(P + R)\n\n
Rollout 4
10 / 2000 = 0.005. Similarly, for C
Rollout 8
you can have either R followed by G or G followed by R. And the lengths of each block are
Rollout 7
must be integer, so N must be a multiple of 12. Just like part (a).

Layer 36

GATE_PROJ

Top 16 Positive Activations
Rollout 8
)! ) * (5! /k! ) ]\n\nThen, probability is [240 + Sum
Rollout 5
, let me check the calculations again step by step. Probably, I made an algebraic mistake.\n\nStarting
Rollout 7
't be. So there must be a mistake here. Let me verify step by step.\n\nWe derived:\n\n
Rollout 9
) with base EM.\n\nSo let's calculate coordinates.\n\nGiven:\n\nE is (10,0)\n
Rollout 8
30240.\n\nSimplify this fraction.\n\nFirst, divide numerator and denominator by 24
Rollout 7
equation with x and N. Let's compute that.\n\nFirst expand:\n\n75*(2N/3
Rollout 1
achieve this for all x, x' in S?\n\nSuppose that all points in S lie on a
Rollout 0
instead of 20!, take a simpler case. Let me check with n = 6. Then
Rollout 7
need to model the original groups and the new groups.\n\nLet me denote variables:\n\nOriginal scenario:\n\n- Number
Rollout 4
\n\nHmm, let's compute these step by step.\n\nFirst, compute the coefficients for ΔC_eq.\n\n
Rollout 0
the exponents of primes in 20!:\n\nFor prime 2: floor(20/
Rollout 1
k such that this holds for all x in S.\n\nBut maybe we can see this as a functional equation
Rollout 8
0P5\n\nAlternatively, compute numerator and denominator.\n\nFirst, compute denominator: 10P5
Rollout 3
22)) Hmm, okay. Let me calculate.\n\nFirst, 8 * 8.31
Rollout 1
all x in S.\n\nBut what does that imply?\n\nLet me set x and x' in S.
Rollout 2
* (-1)^{997} mod8.\n\nFirst, compute (-1)^{99
Top 16 Negative Activations
Rollout 6
's theorem equivalence between \( \textbf{P}/\text{poly} \) and sparse or
Rollout 6
}$ . Show that there exists a sparse set $S_L$ and a deterministic polynomial time TM
Rollout 3
68*0.5 +0.5^2=28,224 +1
Rollout 6
bb{N}$ , a sequence of strings $\{\alpha_n\}_{n \in \math
Rollout 6
theorem equivalence between \( \textbf{P}/\text{poly} \) and sparse oracles
Rollout 5
left-hand side:\n\ny^2 - 125y + 100y -
Rollout 3
168*0.5 +0.5^2=28,224 +
Rollout 6
,1\}^{p(n)}$ , and a deterministic polynomial time Turing Machine $M$
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a trapez
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when $9 \
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number, write it
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants

UP_PROJ

Top 16 Positive Activations
Rollout 6
doesn't work. Another idea is to use pairwise hash functions or some indexing scheme. Wait, but we
Rollout 2
1000 quickly converges due to multiplicative factors. So the answer is109.
Rollout 0
20!.\n\nTo compute the exponents of primes in 20!:\n\nFor prime 2:
Rollout 0
in N's prime factorization), which requires multiplicativity. For each prime p | N, assign all
Rollout 0
and 144. How is it multiplicative, related to the number of prime factors?\n\nYes
Rollout 6
't work. Another idea is to use pairwise hash functions or some indexing scheme. Wait, but we need
Rollout 5
\) and the longer base be \( b + 100 \). The midline of the
Rollout 0
exponents), it depends. As with N! for N>1.\n\nWait, for N >=2
Rollout 7
with \( N \) participants (\( N < 40 \)), the pass mark is fixed at
Rollout 2
when you have multiple terms multiplied, perhaps with some factors of 1000. If the product
Rollout 2
\nTo find the remainder when \(9 \times 99 \times 999 \times
Rollout 6
1/k}, assuming m is a perfect kth power. Otherwise, no solution. Hence, for each
Rollout 3
constant (\(8.314 \, \text{J/(mol·K)}\)),\n
Rollout 9
\nSquare \(AIME\) has sides of length 10 units. Isosceles triangle \(
Rollout 0
of primes, but the product of primes raised to various powers. Wait, so maybe I need to rethink
Rollout 0
prime factorization of 20!.\n\nTo compute the exponents of primes in 20!:\n\n
Top 16 Negative Activations
Rollout 9
E=(10,0), M=(10,10), G=(10 - h
Rollout 9
0,10), and E at (10,0). That's consistent.\n\nNow triangle G
Rollout 9
(10,0)\nM is (10,10)\nG is (10 -
Rollout 9
5), (10,0), (10,10). The overlapping area between this triangle
Rollout 9
E (10,0), M (10,10). So the altitude is indeed h
Rollout 9
coordinates:\n\nA(0,0), I(0,10), M(10,1
Rollout 9
to (10,0) to (10,10). The area of this triangle can
Rollout 9
triangle spans from (10 - h,5) to (10,0) and (1
Rollout 9
EM is the vertical line from (10,0) to (10,10), the
Rollout 9
(10,10), E(10,0). Wait, wait no: in the
Rollout 8
0, all green, and k=5, all red, which are already covered in Case 1
Rollout 9
A at (0,0), I at (0,10), M at (10,
Rollout 1
||y||2 = d2 / 2.\n\nThus:\n\n-2⟨x, y〉 +
Rollout 5
75 = 2b + 150 \implies b = 75\n\
Rollout 8
2. Adding Case 1 (all red, all green), total 10 color sequences.\n\nTherefore
Rollout 9
to (10,0) and (10,10). So these are straight lines.

DOWN_PROJ

Top 16 Positive Activations
Rollout 9
= 100 => y = 5.\n\nTherefore, the y-coordinate of point G must be
Rollout 8
probability that Kathy will be happy is $\frac{m}{n}$ , where $m$ and $
Rollout 3
·mol). Then, dividing by M in kg/mol, the mol unit cancels, then kg in
Rollout 2
1000=2^3 *5^3.\n\nTherefore, if in the product there
Rollout 7
/3.\n\nBut x= N/12.\n\nSo substituting:\n\n(N/12)(A
Rollout 4
, their product is 1990*2985. Let's compute that.\n\n1
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number, write it
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a trapez
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when $9 \
Top 16 Negative Activations
Rollout 1
||2. Therefore:\n\nd2 = (d2)/2 + (d2)/2 -
Rollout 3
2)^2 = 170^2 - 4*170 + 4 =
Rollout 6
, but here the advice is fixed for all inputs of length n, and the machine runs in poly-time
Rollout 8
actually, for k=2, is the number 1200 per sequence? Wait, no
Rollout 1
circular.\n\nAlternatively, setting t = ||x|| squared, you get equations:\n\n-⟨x, y
Rollout 6
allows L to be reduced. So probably the intended answer is to cite/to use this theorem.\n\nHence
Rollout 8
10! / (10-5)! = 10 * 9 * 8
Rollout 3
is wrong, but maybe the problem's answer key uses 298 K.\n\nAlternatively, let's
Rollout 2
8, and 2^k for k ≥3 is 0 mod8. Thus, for
Rollout 3
.5)^2=168^2 + 2*168*0.5
Rollout 1
', x - x'⟩ = ⟨x, x⟩ - 2⟨x, x'⟩
Rollout 8
examples are written in terms of colors.\n\nHence, likely, we can assume that sequences are considered the
Rollout 0
where k is primes. But, since prime factors with exponents: Wait, but each prime is either
Rollout 3
, perhaps the answer expects 170 m/s as an approximate answer. But some precision is required
Rollout 5
5. Compute the RMS: sqrt[(752 + 1752)/2] =
Rollout 2
1= -1 mod1000 fork>=3, because10^3=10

Layer 37

GATE_PROJ

Top 16 Positive Activations
Rollout 5
5). Is that right?\n\nWait, let me check the calculations again step by step. Probably, I
Rollout 3
so I need to be careful here. Let me check.\n\nI recall the root mean square (rms)
Rollout 8
. Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We
Rollout 5
iscalculation in the algebra step. Let me verify quickly. \n\nAlternatively, perhaps it is quicker to
Rollout 3
(Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the
Rollout 4
15 )^2\n\nHmm, let's compute these step by step.\n\nFirst, compute the coefficients
Rollout 7
equation with x and N. Let's compute that.\n\nFirst expand:\n\n75*(2N/3
Rollout 2
by $1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem
Rollout 6
$L$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work
Rollout 9
5}\n\nWait, that seems correct.\n\nLet me check this.\n\nGiven that the area is 10
Rollout 7
us an equation with x and N. Let's compute that.\n\nFirst expand:\n\n75*(2N
Rollout 5
, let me check the calculations again step by step. Probably, I made an algebraic mistake.\n\nStarting
Rollout 5
miscalculation in the algebra step. Let me verify quickly. \n\nAlternatively, perhaps it is quicker
Rollout 8
6=30240. That's correct.\n\nAlternatively, another way to think about it:
Rollout 1
$ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show
Rollout 4
1200. Yep, that's right. So, C_eq is 120
Top 16 Negative Activations
Rollout 8
of ways to choose r red cards and arrange them) * (number of ways to choose 5−
Rollout 4
* Δa )^2 + ( (a2 / (a + b)^2 ) * Δ
Rollout 8
ways to arrange the cards is \(2 \times P(5, k) \times P(5
Rollout 8
color sequence starting with red, you first choose and arrange r red cards, then choose and arrange 5
Rollout 4
( (b2 / (a + b)^2 ) * Δa )^2 + ( (
Rollout 0
. The number of coprime pairs a < b where a*b=720 is 4
Rollout 8
5−r:\n\n- The number of permutations is P(5, r) * P(5,
Rollout 1
the vectors { sqrt(2)/d(x - y) : x ∈ S } are orthonormal
Rollout 0
two choices. Hence no. of pairs (a,b), ordered is 2^8=25
Rollout 6
where i is a bit position in α_n, but then S_L contains these pairs (n,i)
Rollout 6
since it's known in complexity theory that P/poly is equivalent to having a polynomial-time machine with a
Rollout 0
, then exactly half of those subsets gives a*a less than N! , that is, a < sqrt
Rollout 0
that each coprime pair (a,b) with a*b = N is equivalent to choosing a subset
Rollout 0
((a, b)\) and \((b, a)\) are distinct. To find pairs where
Rollout 1
From the sphere condition, ||x - t/2||2 = (d2)/2.\n\nExp
Rollout 7
3.\n\nAfter increase, promoted =P +x where x= original repeaters>=60.\n\nSimilarly

UP_PROJ

Top 16 Positive Activations
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a trapez
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when $9 \
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number, write it
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $A
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say
Top 16 Negative Activations
Rollout 9
be equal. Therefore, the triangle has two equal sides (GE and GM) with base EM.\n\nSo
Rollout 4
C1 + C2)) ]\n\nTherefore, the relative error in C_eq is ΔC_eq / C
Rollout 5
, the ratio of the areas depends on the squares of the lengths involved. The formula for the length of
Rollout 5
) * h/2. Then, the ratio of upper to lower area is (b + 2
Rollout 9
polygon formed is a trapezoid with two vertices on the left edge and two vertices on the right
Rollout 9
must be equal. Therefore, the triangle has two equal sides (GE and GM) with base EM.\n\n
Rollout 1
2, x3 form an equilateral triangle with side length d. Suppose their positions are x1,
Rollout 8
1.\n\nSo happy sequences: the two all red, two all green, and the mixed.\n\nBut wait
Rollout 7
60, average B.\n\nSo:\n\nTotal score of promoted-from-repeaters: A * x.\n\nTotal
Rollout 1
, take three points in R^3 with equal pairwise distances d (i.e., an equilateral triangle
Rollout 4
y^b * z^c, then the relative error in Q is sqrt( (aΔx
Rollout 5
2. Then, the ratio of upper to lower area is (b + 25)/(b +
Rollout 8
Similarly all green: 5!. So total 2*5! = 240.\n\n-
Rollout 6
n. If k is a constant, then the total number for each n would be k*p(n).
Rollout 5
bases a and x and height k. The ratio of heights is k/h. The problem is to find
Rollout 9
. Since the triangle is isosceles with base EM, the apex is G. And the altitude

DOWN_PROJ

Top 16 Positive Activations
Rollout 2
8.\n\nTherefore, equation becomes:\n\n(5m +5)≡5 mod8\n\nSubtract
Rollout 6
textbf{NP} \).\n\n**Answer:** \n\( \textbf{P} = \text
Rollout 1
necessarily?\n\nWait, the problem doesn't state anything on how big the set S is. Given that in
Rollout 1
0.\n\nSo another way to think of this is that the set {x - y : x ∈
Rollout 4
approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of the energy stored
Rollout 9
0 -50/25=(1/5)x +10 -2= (1/
Rollout 6
\) is unique per length.\n\n**Answer:** \nThe sparse set \( S_L \) contains encoded
Rollout 6
\textbf{NP} \).\n\n**Answer:** \n\( \textbf{P} = \
Rollout 4
respect to a is (b*(a + b) - a*b)/(a + b)^2)
Rollout 4
.8412\n\nThus E_max≈15,195.84e-
Rollout 8
) * 5!/(r! ) / (5 - (5 -r))! ; wait
Rollout 7
2/3)N.\n\nSo N must be divisible by 3. Therefore, N is a multiple
Rollout 4
the first term is [0.003]^2. But wait:\n\nFirst term inside the formula
Rollout 1
x'⟩)/2.\n\nSo this gives a condition on y. So each pair x, x'
Rollout 3
169} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon gas is \
Rollout 1
0.\n\nSo another way to think of this is that the set {x - y : x ∈ S
Top 16 Negative Activations
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a trapez
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when $9 \
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number, write it
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides
Rollout 2
before, 10^3=1000≡0 mod125, so
Rollout 1
x, x'⟩ + ⟨x', x'⟩ = d2.\n\nHence, from this
Rollout 5
175 from the midline area ratio 2:3.\n\n2. Remembered that the line
Rollout 9
0/h ) ] dx\n\n= ��[0 to10] [5/h x +1
Rollout 1
$ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to
Rollout 3
Radon is much heavier, 222 g/mol vs 28 g/mol, much slower

Layer 38

GATE_PROJ

Top 16 Positive Activations
Rollout 5
81 is the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n<|im_start|>answer
Rollout 0
8-1}=128.\n\n**Final Answer**\n\boxed{128}\n\n<|im_start|>answer
Rollout 1
}}" or as per problem statement.\n\n**[Final Answer]**\n\boxed{\text{Such a point
Rollout 3
{169} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon gas is
Rollout 9
being h=25.\n\nPerfect.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>answer\n
Rollout 8
26= 157.\n\n**Final Answer**\n\boxed{157}\n\n<|im_start|>answer
Rollout 3
{169} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon gas is
Rollout 2
is \boxed{109}\n\n**Final Answer**\n\boxed{109}\n\n<|im_start|>answer
Rollout 4
is approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of the energy
Rollout 3
answer 169 m/s.\n\n**Final Answer**\n\boxed{170} m/s
Rollout 7
crucial point. Let's read the original problem again:\n\n"due to an error in the wording of a
Rollout 2
9*10 +9, wait, perhaps 9999=9*111
Rollout 7
, part(b) has no solutions.\n\n**Final Answer**\n\n(a) \boxed{12}, \
Rollout 7
.\n\nWait, but actually, the problem says: "due to an error in the wording of a question
Rollout 9
: in the problem statement, the square is called AIME. If moving around square AIME, probably
Rollout 9
, no, in that labeling, if AIME is the square, then the sides are AI, IM
Top 16 Negative Activations
Rollout 4
{eq}} \) is:\n\n\[\n\frac{\Delta C_{\text{eq}}}{
Rollout 3
assistant.\n<|im_start|>user\nCompute the mean molecular speed v in the heavy gas radon (Rn)
Rollout 0
problem.\n\nFirst, a rational number between 0 and 1 can be written as a fraction \(\
Rollout 4
the relative error in C_eq is ΔC_eq / C_eq = sqrt( [ (C2 /
Rollout 0
it says that the original rational number between 0 and 1 is written in reduced form, and you
Rollout 0
<|im_start|>answer\nGiven a rational number between 0 and 1, we need to write it as a
Rollout 4
frac{\Delta C_{\text{eq}}}{C_{\text{eq}}} = \frac
Rollout 0
First, a rational number between 0 and 1 can be written as a fraction \(\frac{
Rollout 7
0). Their new scores are original+5 (<=64). Average should be47. So
Rollout 3
al., there is an example: Calculate the average speed of N2 molecules at 25°C.
Rollout 8
of greens is 5 - k from 4 to 1), there are two color sequences: R
Rollout 4
energy would then be the square root of the sum of the squares of the relative errors of each variable,
Rollout 7
average=A=A_old +5, between65 and69.\n\nTotal score of new promoted=A*x
Rollout 5
) \implies 3b + 75 = 2b + 150 \
Rollout 0
fractions a/b where 0 < a/b <1 and a*b=20!. Hence, equivalent
Rollout 1
- x'⟩ = ⟨x, x⟩ - 2⟨x, x'⟩ + ⟨

UP_PROJ

Top 16 Positive Activations
Rollout 1
)/2 = (3d2 / 2)/2 = 3d2/4.\n\n〈
Rollout 1
||2 + ||x'||2 - d2)/2.\n\nBut perhaps another way is expressing this by
Rollout 9
So indeed, (0,10 -50/h) is between (0,5) and
Rollout 1
(||x||2 - ||x'||2)/2.\n\nBut if we assume that ⟨x -
Rollout 6
(log n + log p(n)) = O(log n) (since p(n) is poly(n),
Rollout 5
1: h = (125H)/(75 + x). Plug into equation 2
Rollout 1
(||t||2 - 2d2)/4.\n\nTherefore, for all x in S,\n\n
Rollout 5
first equation: h = (125H)/(75 + x)\nFrom the second equation:
Rollout 8
: 5!. So total 2*5! = 240.\n\n- For the mixed
Rollout 8
the number of card sequences is 2 * [P(5, k) * P(5,
Rollout 1
2). Since 2r = 2d / sqrt(2) = d sqrt(2)
Rollout 5
2 = (2b + 100)/2 = b + 50. But the
Rollout 1
x1||2 - ||x2||2)/2.\n\nSimilarly, using the distance between x1
Rollout 1
, x⟩ - ⟨x', x'⟩)/2.\n\nBut also, since ||x - x
Rollout 5
175)/2 * (H - h) = 62.5H.\n\nSo
Rollout 3
s2·mol) divided by (kg/mol) is (m2/s2). Square root is
Top 16 Negative Activations
Rollout 9
area common to triangle \(GEM\) and square \(AIME\) is 80 square units.
Rollout 5
181.25\n\]\nThe greatest integer not exceeding 181.25
Rollout 8
7440\).\n\nThe probability is \(\frac{7440}{30
Rollout 9
\boxed{25}\n\n<|im_start|>answer\nSquare \(AIME\) has sides of length 10
Rollout 3
{m/s}\).\n\n**Final Answer:**\n\[\n\boxed{169} \,
Rollout 8
one block of red and one block of green.\n\n1. **Case 1: All red or all
Rollout 7
), \(\boxed{24}\), and \(\boxed{36}\).\n\n(b) No
Rollout 8
40}{30240} = \frac{31}{126}\).
Rollout 6
{bad-angel}} \) unless \( \textbf{NP} = \textbf{P
Rollout 1
} y \text{ exists as described}}\n\]<|im_end|>
Rollout 9
\) has sides of length 10 units. Isosceles triangle \(GEM\) has base
Rollout 8
\[\n\boxed{157}\n\]<|im_end|>
Rollout 6
} \subseteq \textbf{P}_{\text{bad-angel}} \) by setting \(
Rollout 6
\neq \textbf{P}_{\text{bad-angel}} \) unless \( \
Rollout 0
, the number of rational numbers between 0 and 1 for which the product of the numerator and denominator
Rollout 7
oted $47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve

DOWN_PROJ

Top 16 Positive Activations
Rollout 7
5-point increase, every participant's score is increased by 5. Therefore, the new total score
Rollout 1
/ 2)/2 = 3d2/4.\n\n〈y, y〉 = ||y
Rollout 8
from 1 to 4 reds and 4 to 1 greens).\n\nWait, so for each
Rollout 6
can query whether $s \in S$ and get the correct answer in return in constant time
Rollout 7
71P + 5P = 76P, and for repeaters, their total becomes
Rollout 4
by 5025:\n\nFirst, 5025 * 1200 =
Rollout 1
.\n\nHence, from this, we can solve for ⟨x, x'⟩:\n\n⟨x,
Rollout 7
and the average of the non-promoted (repeaters) becomes 59.\n\nPart (a)
Rollout 4
:\n\nFirst, 5025 * 1200 = 6,030
Rollout 5
125\n\nFactor the second term:\n\nLeft side = (x + 175) -
Rollout 5
, and area (x + 175)/2 * (H - h) = 6
Rollout 7
9x\n\nMove terms:\n\n76P -79P =79x - (A +
Rollout 4
00, since 5025*1000=5,025,
Rollout 7
all scores are increased by 5. After this increase, the average of the promoted participants becomes 7
Rollout 3
multiplication.\n\n66.512 * 300 = 19,953
Rollout 9
10 -0=10) and height=10 (the distance between x=0
Top 16 Negative Activations
Rollout 8
2 * [P(5, k) * P(5, 5 - k)] where k
Rollout 5
0)] / 2 = [150 ± sqrt(72500)] /
Rollout 8
example, if k=2, the number of ordered hands would be:\n\nFirst choose 2 red cards
Rollout 8
10 different cards, hence the number of ordered sequences is 10P5. If they
Rollout 8
). So 4*3=12 possible ordered sequences.\n\nBut applying our previous approach: All red
Rollout 8
5 green. Then, the total number of possible ordered hands (sequences) is 10 *
Rollout 9
,50/h) and (0,10 -50/h). The overlapping area between the
Rollout 9
), h>10. The triangle is isosceles, so when G is outside, the
Rollout 8
(choosing 5−r greens and permuting them). Thus, for a particular r and
Rollout 0
2^8=256, each ordered pairs. Taking half because a < b hence
Rollout 3
depending on source of problem, maybe requires two sig figs. 222 is three, so
Rollout 7
we have a group of x = N/12 students who had original average 62. So
Rollout 8
of ordered hands is P(5,k) * P(5,5 -k). For example,
Rollout 3
8.314 \, \text{J/(mol·K)}\)),\n- \( T
Rollout 8
positions. This is equal to P(5,2) * P(5,3), where P
Rollout 7
2N/12 +54N/4 =62N/12 +16

Layer 39

GATE_PROJ

Top 16 Positive Activations
Rollout 3
calculate the mean molecular speed of radon gas. Hmm, first, I remember there's a formula for
Rollout 4
00 ± 0.02 V. Hmm, alright, let's start by recalling how to
Rollout 2
, is divided by 1000. Hmm, okay, let's unpack this step by step
Rollout 2
-3 \equiv 5 \mod 8\n \]\n\n2. **Modulo 1
Rollout 2
\]\n\n2. **Modulo 125 Calculation:**\n - Each term \(10
Rollout 2
\equiv 5 \mod 8\n \]\n\n2. **Modulo 125
Rollout 2
we can exploit when multiplying numbers modulo 1000.\n\nAlternatively, since 1000
Rollout 4
_{1}=2000 \pm 10 \mathrm{pF}$ and $C
Rollout 2
^k - 1\), which modulo 8 is \(2^k - 1\).\n
Rollout 2
}_{\text{999 9's}}$ is divided by $1000$
Rollout 9
triangle \(GEM\) and square \(AIME\) is 80 square units. We need to
Rollout 2
8\).\n - The product modulo 8 is:\n \[\n (2^1
Rollout 2
8.\n\nSince 10 �� 2 mod 8, so 10^k
Rollout 8
times 5 \times 4 \times 3) = 2400\)\n
Rollout 3
the mean molecular speed v in the heavy gas radon (Rn) in m/s\n<|im_start|>assistant
Rollout 2
der Theorem.\n\n1. **Modulo 8 Calculation:**\n - Each term in the product
Top 16 Negative Activations
Rollout 3
this question is ambiguous. Nevertheless, likely answer \boxed{170} m/s. Alternatively,
Rollout 3
the mean molecular speed of radon gas is \boxed{170} m/s if we round
Rollout 3
m/s. Thus, present answer boxed as \boxed{169 \text{ m/s}}
Rollout 3
50)=161.4, as 161^2=25,9
Rollout 3
K (unless stated otherwise), perhaps answer around \boxed{169} m/s. But need
Rollout 2
125 mod8: 125 /8= 15*8=12
Rollout 3
, the mean molecular speed of radon gas is \boxed{170} m/s if we
Rollout 9
=25. Thus, the answer is \boxed{25} . Seems done.\n\nThat's
Rollout 3
m/s.\n\nTherefore the valid boxed answer is \boxed{169}.\n\nBut initially I thought
Rollout 8
0. Let's see: 240×30 =7200, 7
Rollout 3
, this question is ambiguous. Nevertheless, likely answer \boxed{170} m/s. Alternatively
Rollout 7
0. Therefore, x can vary from 0 up to N/3. Since x= N/
Rollout 3
8,408). Let's see.\n\n168^2 = (170 -
Rollout 3
square root method:\n\n168.0^2=28,224.0\n\n
Rollout 3
). Let's see.\n\n168^2 = (170 - 2)^2 =
Rollout 3
0, which is incorrect. So correct is \boxed{169} m/s.\n\n**Final

UP_PROJ

Top 16 Positive Activations
Rollout 8
\(240 + 7200 = 7440\).\n\nThe probability is
Rollout 8
2400 + 1200 = 7200\).\n\nAdding both cases
Rollout 4
{eq}}} = \frac{4.327}{1200} \approx
Rollout 8
+ n = 31 + 126 = 157\).\n\n\[\n\
Rollout 8
2400 + 1200 = 7200\).\n\nAdding both
Rollout 5
. The length \( x \) is given by:\n\[\nx = \sqrt{\frac{a
Rollout 4
.96 + 5.76} = \sqrt{18.72} \
Rollout 4
.16 \cdot 15)^2} = \sqrt{3.6^2 +
Rollout 5
0 ± 10*sqrt(725)] / 2 = 75 ±
Rollout 4
Percentage error} = 0.008775 \times 100 \approx
Rollout 4
16 \cdot 15)^2} = \sqrt{3.6^2 +
Rollout 4
^2 + 2.4^2} = \sqrt{12.96 +
Rollout 4
0.16 \cdot 15)^2} = \sqrt{3.6^2
Rollout 5
= 2(b + 75) \implies 3b + 75 =
Rollout 8
8 \times 7 \times 6 = 30240\).\n\nNext,
Rollout 8
total number of happy sequences is \(240 + 7200 = 744
Top 16 Negative Activations
Rollout 8
, examples include RRRRR and GGGGG.\n\nSo actually, if all red cards in the layout
Rollout 9
0,0) lies within the square? Yes.\n\nWait, but then the area of the triangle in
Rollout 9
), where h is the altitude. Or is the altitude segment inside the triangle? Wait. Wait, in
Rollout 9
with base EM. Since triangle GEM is isosceles, meaning legs GE and GM are equal
Rollout 9
- h, y). But the triangle being isosceles with base EM. Since triangle GEM
Rollout 8
of the cards in a row in a random order."\n\nWait, the problem doesn't specify whether the cards
Rollout 9
the triangle being isosceles with base EM. Since triangle GEM is isosceles,
Rollout 0
fraction as part of between 0 and 1. When written in lowest terms, numerator a and denominator
Rollout 9
so the triangle is completely within the square. Wait, because the square spans from x=0 to x
Rollout 9
. Then the triangle GEM overlaps the square, and their common area is 80 square units.
Rollout 9
so overlap area is 50, as before. So how can we have overlapping area of 8
Rollout 9
triangle is within. But then how come the overlapping area is 80 then? Because that that case
Rollout 9
base of the triangle. Wait, no. Wait, EM is the top side of the square. Wait
Rollout 8
just says 5 red and 5 green cards.\n\nIs each red card identical and each green card identical
Rollout 9
the triangle? Wait. Wait, in an isosceles triangle, the altitude from the apex to
Rollout 9
of the triangle is inside the square (the common area is 80), and the square's right

DOWN_PROJ

Top 16 Positive Activations
Rollout 1
d / sqrt(2) times sqrt(2) from center y, which is exactly d. Wait
Rollout 0
b = 20!\) and \(a < b\).\n\nNow, how do I count such
Rollout 4
sqrt(7.696 x 10^-5). sqrt(7.696
Rollout 8
/30240=744/3024=744÷2
Rollout 9
area between these two lines is the area of trapezoid with bases at x=0 (length
Rollout 1
all be distance d / sqrt(2) times sqrt(2) from center y, which is exactly
Rollout 5
and x with height h has area (75 + x)/2 * h = 62.
Rollout 9
50/h +5/h x -50/h] dx\n\n= ��[0 to1
Rollout 0
answer\nGiven a rational number between 0 and 1, we need to write it as a fraction
Rollout 0
a\) and \(b\), with \(a < b\). Since \(a\) and \(b
Rollout 9
simpler way is to compute the area of the trapezoid formed between these two lines, from x
Rollout 7
:\n\n71P +56(N -P)=66N\n\nExpand:71P +
Rollout 0
8}\n\n<|im_start|>answer\nGiven a rational number between 0 and 1, we need to write it
Rollout 1
x⟩ - ⟨x', t - x'⟩ = 0 ⇒ ⟨x, t⟩ -
Rollout 6
. (This is the Meyer, I think, theorem.) So perhaps this is the intended approach. That
Rollout 1
- ||x2||2 ⇒ ⟨x1 - x2, y⟩ = (||x1
Top 16 Negative Activations
Rollout 3
/337≈0.047. Hence, 168.5 +0
Rollout 4
00)/(10,000,000) = 0.003
Rollout 1
||x_i - y|| = d / sqrt(2).\n\nSo if such a y exists, then
Rollout 4
4*10^6)/(25*10^6)=4/25=0
Rollout 0
19 each have exponent 1 in 20!.\n\nSo exponents are: 18
Rollout 7
<40. 36 is less than 40, so all three are acceptable.\n\nWait
Rollout 8
it's P(5,3) * P(5,2) for the green and red cards
Rollout 8
so sum m +n= 31 +126= 157.\n\n**Final
Rollout 5
Divide both sides by H (assuming H ≠ 0):\n\n(x + 175) -
Rollout 1
}, so r1 + r2 = d sqrt(2) �� 1.414
Rollout 1
^2 - d2) / 2 = 0.\n\nCompare to the above line:\n\nFrom the
Rollout 9
is impossible. Alternatively, if h is greater than10, then the overlap is the portion of the
Rollout 6
angel. But according to the above, P_bad-angel = P, so NP = P_bad-
Rollout 3
figs), T is 298 K (3 sig figs assuming given as 29
Rollout 2
0 mod8. Therefore, m=8n, so x=125*8n +
Rollout 6
here. Given the time constraints, maybe for problem 3, since it's known in complexity theory that

Layer 40

GATE_PROJ

Top 16 Positive Activations
Rollout 4
,150 by 5025:\n\nFirst, 5025 * 1
Rollout 8
divided by 240. Let's see: 240×30 =72
Rollout 8
0 divided by 240. Let's see: 240×30 =7
Rollout 3
(28,436). Let's see, 168^2 = 2
Rollout 4
327 divided by 1200.\n\nDivide by 100: 4
Rollout 4
5025 �� let's divide 6,060,150 /
Rollout 4
/ 5025. \n\nDividing 6,060,150 by
Rollout 2
0^4 mod125 -1). 10^3=0 mod125
Rollout 3
222)) Hmm, okay. Let me calculate.\n\nFirst, 8 * 8.3
Rollout 8
0240.\n\nSimplify this fraction.\n\nFirst, divide numerator and denominator by 240
Rollout 3
12 * 298 = let's compute 66.512 * 3
Rollout 4
7 / 1200. Let me calculate 4.327 divided by 1
Rollout 4
%. Wait, let me check the exact value.\n\nFirst, compute (ΔC_eq / C_eq):\n\n
Rollout 8
1/126.\n\nWait, wait: 7440 ��240:
Rollout 4
15 )^2\n\nHmm, let's compute these step by step.\n\nFirst, compute the coefficients
Rollout 8
.246.\n\nThus, the answer would be 31 +126=15
Top 16 Negative Activations
Rollout 1
a Hilbert space such that the images are orthonormal.\n\nBut here, we're dealing within the
Rollout 1
'⟩ = (2||x||2 - d2)/2.\n\nBut we don't know if
Rollout 1
x||2 + ⟨y, y⟩ - d2 / 2.\n\nBut if ||x||
Rollout 1
S\right\}\] \nis an orthonormal system of vectors in $ \mathcal{
Rollout 1
bert space, one can have uncountable orthonormal systems.\n - The intersection of infinite closed
Rollout 1
\n\[ \left\{\frac{\sqrt{2}}{d}(x\minus{}y):
Rollout 0
Given a rational number, write it as a fraction in lowest terms and calculate the product of the resulting numerator
Rollout 1
2y⟩ being equal to ⟨x, x⟩ + k for all x ∈ S.\n\nWait
Rollout 0
by 2, excluding the case \(a = b\) (which is impossible since 20!
Rollout 1
space, one can have uncountable orthonormal systems.\n - The intersection of infinite closed convex
Rollout 1
- y) : x ∈ S } are orthonormal. Hence, they can be considered as an
Rollout 1
-empty if and only if the distance between x1 and x2 is ≤ 2r. Here,
Rollout 0
of the total ordered pairs (excluding the a = b case, which does not exist here). Therefore,
Rollout 1
)/2 - 2⟨x - y, x' - y⟩.\n\nTherefore, this simplifies
Rollout 0
user\nGiven a rational number, write it as a fraction in lowest terms and calculate the product of the
Rollout 4
the uncertainty in \( C_{\text{eq}} \) using error propagation. The partial derivatives are

UP_PROJ

Top 16 Positive Activations
Rollout 0
(even), 2 (7) (even?), no 7 has exponent 2? Wait for
Rollout 6
, not the other way.\n\nWait, but the problem doesn't say that k is given as part of
Rollout 8
* 63=2*7*9. 31 and 126 are cop
Rollout 9
,0). Wait, wait no: in the problem statement, the square is called AIME. If
Rollout 6
how does S know about k? Because in the problem statement, given S_1,...,S_k
Rollout 1
- d2 - 2c + d2)/2 = 0 ⇒ 0 = 0
Rollout 0
the total by 2.\n\nWait, but the problem says "how many rational numbers between 0 and
Rollout 6
_bad-angel unless NP = P.\n\nBut the problem says "Is NP = P_bad-angel?
Rollout 7
problem states N<40. 36 is less than 40, so all three are
Rollout 7
47)there is no solution. But the problem says "Find all possible values of N", so
Rollout 7
(a), they change to different averages?\n\nBut problem says: "due to an error... all scores
Rollout 9
5)/(10 - (10 - h)) = 5/h.\n\nTherefore, equation is y
Rollout 7
b) impossible, no solutions. But language problems say "Find all possible values ofN". If "
Rollout 7
's a critical point.\n\nWait, but reading the problem again: "due to an error in the wording
Rollout 7
, we need to say no solution. But the problem probably expects some answer, possibly also N=1
Rollout 1
is not necessarily separable here.\n\nWait, the problem allows H to be non-separable. But even
Top 16 Negative Activations
Rollout 1
y. So the main idea is to find a point y such that all x in S are at distance
Rollout 1
2).\n\nSo, how do we show that such a y exists?\n\nIn finite dimensions, if I have
Rollout 4
about the percentage error in the energy, I need to figure out how the uncertainties in C1, C
Rollout 4
the energy stored in a capacitor is (1/2) * C_eq * V2. Since the
Rollout 5
, if there is a segment parallel to the bases that divides the trapezoid into two regions of
Rollout 1
d.\n\nNow, the problem wants me to find a point y such that when I take (sqrt(
Rollout 5
the lengths involved. The formula for the length of a line segment parallel to the bases that creates a certain
Rollout 1
orthonormal.\n\nSo the key is to find a point y such that all x in S lie on
Rollout 1
||x - y||2 = d2 / 2.\n\nExpanding, that gives:\n\n⟨x
Rollout 1
Therefore, ⟨x - y, x' - y⟩ = 0.\n\nSo another way to think
Rollout 1
thonormal.\n\nSo the key is to find a point y such that all x in S lie on a
Rollout 4
given by \( E = \frac{1}{2} C_{\text{eq}} V^
Rollout 1
.\n\nTherefore, the key is to show that such a point y exists, which is equidistant to
Rollout 1
pairwise orthogonal.\n\nSo the question reduces to showing such a y exists given that all pairwise distances in S are
Rollout 1
. So the main idea is to find a point y such that all x in S are at distance d
Rollout 4
the energy. The energy stored is (1/2) * C_eq * V^2. So

DOWN_PROJ

Top 16 Positive Activations
Rollout 1
1 + r2 >= l and |r1 - r2| <= l.\n\nIn our case,
Rollout 9
5)/h.\n\nThe equation is y -0 = (-5)/h (x -10).
Rollout 1
if we have a family of closed convex sets where every finite intersection is non-empty, then the whole intersection
Rollout 0
2) + floor(20/4) + floor(20/8) + floor(
Rollout 8
can be simplified. Since 31 is prime. 126 divides by 2,
Rollout 0
product of two coprime numbers, \(a\) and \(b\), with \(a < b
Rollout 3
"mean molecular speed"—is that the same as the root mean square speed, or is it a different
Rollout 4
*0.8 +597*0.0004=477.6
Rollout 1
thonormal vectors must have that ⟨ x - y, x' - y �� = 0.\n\n
Rollout 0
the resulting numerator and denominator. For how many rational numbers between 0 and 1 will $20
Rollout 4
differentiation. Taking the natural logarithm of C_eq: ln C_eq = ln C1 + ln C
Rollout 1
x', the vector x - y and x' - y must be orthogonal. That suggests that the inner
Rollout 0
prime factorization of 20!.\n\nTo compute the exponents of primes in 20!:\n\n
Rollout 2
5 and8 are coprime, m must be ��0 mod8. Therefore, m
Rollout 1
x', ⟨√2/d (x - y), ��2/d (x' - y)
Rollout 3
/2.828≈168, which is about right. So, the number looks
Top 16 Negative Activations
Rollout 4
V separately.\n\nStarting with C1: 2000 ± 10 pF. The
Rollout 4
025. \n\nDividing 6,060,150 by 50
Rollout 4
25 �� let's divide 6,060,150 / 50
Rollout 5
100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve
Rollout 1
, t - x⟩ = C,\n\n⟨x', t - x'⟩ = C.\n\nSubtract
Rollout 0
be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how many
Rollout 2
000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here: I
Rollout 4
To maximize this expression, ideally if C1 and C2 are maximum.\n\nWait, but because (C
Rollout 9
triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry problem
Rollout 5
12500 = 125y\n\nSimplify:\n\ny^2 - 2
Rollout 3
Let me check a bit more accurately.\n\n28,408.12 is the number.
Rollout 4
Let me compute it.\n\nWell, 0.000077 is 7.7
Rollout 8
m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to find
Rollout 7
0 but <65). Their maximum possible original score would have been64. Hence, maximum average
Rollout 4
.\n\nStarting with C1: 2000 ± 10 pF. The relative error
Rollout 2
0.\n\nNow, let's approach this modulo 8 and modulo 125.\n\nFirst, modulo

Layer 41

GATE_PROJ

Top 16 Positive Activations
Rollout 3
8 K (25°C) yields ~168.5 m/s. Therefore, since the
Rollout 3
and rounded to three sig figs is 169 m/s. Thus, present answer boxed as
Rollout 3
compute", so it's expected to compute.\n\nAlternatively, perhaps check if there is an established average speed for
Rollout 3
precision needed.\n\nAlternatively, maybe the original question assumes a specific temperature but forgot to mention. If the user
Rollout 3
and calculator computations, perhaps the answer expects 170 m/s as an approximate answer. But some
Rollout 3
98 K (25°C) yields ~168.5 m/s. Therefore, since
Rollout 3
. Since the user hasn't specified temperature, perhaps there's a standard temperature to assume. Let me check
Rollout 3
different.\n\nBut given that the user hasn't specified the temperature, this seems ambiguous. However, in many
Rollout 3
is rounded to three significant figures, giving \(169 \, \text{m/s}\).\n\n
Rollout 3
around 298 K. Maybe. Alternatively, as room temperature sometimes is taken as 30
Rollout 3
's expected to compute.\n\nAlternatively, perhaps check if there is an established average speed for radon gas at
Rollout 3
, given that no temperature is given, perhaps they just want the formula, but I don't think so
Rollout 3
, if forced to choose, answer approximately 170 m/s. But given compute precisely, it
Rollout 3
m/s. But if the question hasn't specified the temperature, this can vary.\n\nAlternatively, given that
Rollout 3
we need to assume a standard temperature.\n\nAlternatively, maybe they expect us to recall the molar mass of
Rollout 3
the absence of a specified temperature, perhaps we need to answer that the temperature is required, but the problem
Top 16 Negative Activations
Rollout 3
Let's see.\n\n168^2 = (170 - 2)^2 =
Rollout 8
0×30 =7200, 7440-7200=
Rollout 4
,030,000, since 5025*1000=
Rollout 4
24=14,328 and 597*0.8004
Rollout 8
0. Let's see: 240×30 =7200, 7
Rollout 8
by 240. Let's see: 240×30 =720
Rollout 4
0 by 5025:\n\nFirst, 5025 * 1200
Rollout 4
6,000,000 and 2000*15=30
Rollout 2
. 891 mod125: 125*7=875.
Rollout 2
125-120=5; 125≡5 mod8.\n\nSimilarly
Rollout 2
=1000 mod125. 1000 divided by125 is
Rollout 4
5,920,250. 5,940,150 -
Rollout 8
40×30 =7200, 7440-7200
Rollout 2
1 mod125: 125*7=875. 891
Rollout 8
's see: 240×30 =7200, 7440
Rollout 4
=30,150. Then, 30,150 /502

UP_PROJ

Top 16 Positive Activations
Rollout 0
but their exponents are 1, which is odd. Actually, wait nvm, the exponents
Rollout 9
). The altitude is h, the horizontal distance from G to EM. So h is the length we need
Rollout 2
7 terms) is-1 mod1000. So multiplying 891 by (-1
Rollout 0
:4,7:2, which are all even? Wait, 18,8,4
Rollout 4
, so 0.88% is already two significant figures. Alternatively, if written as 0
Rollout 1
= sqrt(2) * ||x - y|| / sqrt{1 - a^, but this
Rollout 9
care. So GEM is an isosceles triangle with base EM, so EM is from (
Rollout 0
128\), since each pair is counted twice. But actually, since we are considering ordered pairs
Rollout 0
exponent of 7 is 2, which is even. Wait a second. Then primes 11
Rollout 1
- y) : x ∈ S } are orthonormal. Hence, they can be considered as an
Rollout 1
the vectors from the centroid to the vertices are orthonormal. Wait, not exactly, but in the
Rollout 0
each unordered pair {a, b} is counted twice in the 256 ordered pairs.\n\nBut
Rollout 2
k -1= -1 mod1000 forall k>=3. Therefore, product becomes9
Rollout 1
ormal system. That means each vector has length 1 and any two distinct vectors are orthogonal.\n\nFirst,
Rollout 0
256\), but each pair is counted twice except when \(a = b\). But since
Rollout 0
18 for exponent of 2. Which is even. 3: 8, even.
Top 16 Negative Activations
Rollout 6
time.\n\nSo perhaps the answer is not straightforward, hence why reference to the P/poly theorem. The
Rollout 6
in circles here. Given the time constraints, maybe for problem 3, since it's known in complexity
Rollout 1
only countable ones.\n\nBut the question is probably supposed to allow the System S to be mapped to or
Rollout 6
P/poly.\n\nTherefore, for the first problem, combining multiple sparse sets into one via tagging. For
Rollout 1
the subject.\n\nAlternatively, but, finally, the answer is asserting that such a y exists because he’s
Rollout 6
problem asks to show it, not just cite, perhaps construction:\n\nGiven L ∈ P_angel, then
Rollout 6
Given the time constraints, maybe for problem 3, since it's known in complexity theory that P/p
Rollout 6
'm going in circles here. Given the time constraints, maybe for problem 3, since it's known
Rollout 6
going in circles here. Given the time constraints, maybe for problem 3, since it's known in
Rollout 6
.\n\nSo perhaps the answer is not straightforward, hence why reference to the P/poly theorem. The connection
Rollout 1
such a y exists. Thus, finally, the answer is "\boxed{y \text{ exists as
Rollout 6
circles here. Given the time constraints, maybe for problem 3, since it's known in complexity theory
Rollout 6
polynomial time.\n\nSo perhaps the answer is not straightforward, hence why reference to the P/poly theorem.
Rollout 1
that, after struggling through a number of different avenues, I'll try to formalize the key steps.\n\n
Rollout 1
uses the separability and completeness.\n\nBut maybe an innovative approach is to use the idea of completing the system
Rollout 6
the definition of P/poly.\n\nTherefore, for the first problem, combining multiple sparse sets into one via

DOWN_PROJ

Top 16 Positive Activations
Rollout 4
sqrt(0.000064) is 0.008, and sqrt
Rollout 4
=30,150. Then, 30,150 /502
Rollout 4
0, since 5025*1000=5,025,0
Rollout 4
14,328 and 597*0.8004≈ 5
Rollout 4
=19,900. Then, 19,900 /497
Rollout 4
,000 and 2000*15=30,000 and
Rollout 1
function f(x) = 2⟨x, y⟩ - ⟨x, x⟩ is constant on
Rollout 2
.\n\nHence, the solution is x≡109 mod1000. Therefore, P
Rollout 1
have ⟨x, x'⟩ - ⟨x, y⟩ - ⟨x', y⟩ + ⟨
Rollout 4
6,000,000 and 2000*15=30
Rollout 3
3*66.512:\n\n70*66.512=4,
Rollout 2
*(1000 -1)=891,000 -891=8
Rollout 5
see. I need to solve this problem about a trapezoid with bases differing by 10
Rollout 9
(10, 10), and E is at (10, 0). So the
Rollout 2
0 divided by125 is8, so 1000≡0 mod125
Rollout 1
= ⟨x, x'⟩ - ⟨x, y⟩ - ⟨x', y⟩ + ⟨
Top 16 Negative Activations
Rollout 4
1200. Yep, that's right. So, C_eq is 120
Rollout 9
,0)):\n\nThe slope (m) is (0 -5)/(10 - (10
Rollout 3
8≈168, which is about right. So, the number looks consistent. Therefore, if
Rollout 4
respectively. Then, error terms calculated for each. \n\nAlternatively, perhaps I could use logarithmic differentiation.
Rollout 2
..., up to k=999 (10^999 -1). Therefore, the
Rollout 8
,2) = same 60*20= 1200. Times 2
Rollout 9
0 - (10 - h)) = (-5)/h.\n\nThe equation is y -0 =
Rollout 8
5*4)=2*60*20=2400\n\nk=4:\n\n
Rollout 8
sequences: R GGGG, GGGG R\n\nk = 2: RR GGG,
Rollout 4
, etc., would give the same result as before.\n\nLet me actually compute it this way:\n\nFirst term
Rollout 9
:\n\nThe slope (m) is (0 -5)/(10 - (10 - h))
Rollout 9
E(10,0)):\n\nThe slope (m) is (0 -5)/(10
Rollout 4
formula for percentage error in energy would then be the square root of the sum of the squares of the relative
Rollout 9
10,10).\n\nDistance GM = sqrt[(10 - (10 - h))^2
Rollout 8
00\n\nk=3:\n\nSame as k=2: 2*P(5,3
Rollout 2
109 is odd, which is consistent.\n\nSimilarly, modulus 8 of5 is also odd

Layer 42

GATE_PROJ

Top 16 Positive Activations
Rollout 6
{N} \mapsto \mathbb{N}$ such that for each $n \
Rollout 6
S$ is bounded by $p(n)$ . In other words, $|S^{
Rollout 1
orthogonal.\n\nFirst, let me recall that an orthonormal system requires two things: 1) each
Rollout 6
} \mapsto \mathbb{N}$ , a sequence of strings $\{\alpha_n\
Rollout 6
N} \mapsto \mathbb{N}$ such that for each $n \in
Rollout 6
|S^{=n}| \leq p(n)$ , where $S^{=n}
Rollout 1
S\right\}\] \nis an orthonormal system of vectors in $ \mathcal{
Rollout 8
$m$ and $n$ are relatively prime positive integers. Find $m + n$ .\n<|im_start|>
Rollout 5
into two regions with areas in the ratio 2:3. Then, we need to find x,
Rollout 6
{N} \mapsto \mathbb{N}$ , a sequence of strings $\{\alpha
Rollout 6
$S$ is bounded by $p(n)$ . In other words, $|
Rollout 1
)/d, the resulting set is an orthonormal system. That means each vector has length 1
Rollout 4
00,000) = 0.003\n\nWait, no, actually.
Rollout 6
\{0,1\}^{p(n)}$ , and a deterministic polynomial time Turing Machine
Rollout 6
bb{N} \mapsto \mathbb{N}$ , a sequence of strings $\{\
Rollout 7
5, if someone scores is adjusted beyond100, but the problem doesn't state a maximum score
Top 16 Negative Activations
Rollout 8
), or all reds or all greens.\n\nSo first, we can calculate the number of such happy sequences
Rollout 8
0P5).\n\nLet me try this approach.\n\nFirst, how many color sequences are "happy"?\n\nCase
Rollout 4
9 J=15 nJ.\n\nSimilarly, compute E_max=0.5*120
Rollout 8
P(5,5 -k)]\n\nLet me compute this for k=1 to k=4.\n\n
Rollout 8
30240.\n\nNow, we need to count how many of these sequences have all red cards
Rollout 9
). Connecting (10,0) to (10,10), to (0,1
Rollout 8
First, how many color sequences are "happy"?\n\nCase 1: All reds (RRRRR
Rollout 3
22 kg/mol.\n\nNow, the average speed is sqrt(8RT/(πM)).\n\nBut
Rollout 3
))\n\nLet's calculate the numerator and denominator step by step.\n\n8*R*T = 8 * 8
Rollout 3
.314 * 298\n\nFirst, compute 8*8.314
Rollout 4
2 *5.002\n\nMinimum energy would be with C_eq_min=1194
Rollout 9
lines is the area of trapezoid with bases at x=0 (length 8-2
Rollout 9
25}\n\nWait, that seems correct.\n\nLet me check this.\n\nGiven that the area is 1
Rollout 3
.512≈ let's compute exact:\n\n273 * 66.512
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 1
fix y. Then the conditions we have on y are:\n\n1. For each x in S, ||

UP_PROJ

Top 16 Positive Activations
Rollout 8
31/126, so m=31, n=126.\n\nWait
Rollout 5
the greatest integer not exceeding that is 181. Therefore, the answer is 181
Rollout 8
126, so m=31, n=126.\n\nWait, but let's
Rollout 5
181.25 is 181. Therefore, final answer is 181
Rollout 5
the greatest integer not exceeding this is 181.\n\nWait, this seems straightforward now, but maybe
Rollout 8
26, so m=31, n=126.\n\nWait, but let's confirm
Rollout 9
=25. Thus, the answer is \boxed{25} . Seems done.\n\nThat's
Rollout 5
1. Therefore, the answer is 181. \n\nWait, but let me verify once again
Rollout 5
is 181. Therefore, the answer is 181. \n\nWait, but let
Rollout 4
ounding to two decimal places, 0.88%. However, the question says "percentage error",
Rollout 1
ram law says that ||u + v||2 + ||u - v||2 = 2||
Rollout 4
74%, which rounds to 0.88%. Wait, let me check the exact value.\n\n
Rollout 5
is 181. Therefore, final answer is 181. \n\nI think that's
Rollout 4
, the percentage error is approximately 0.88%, if we round to two decimal places. But
Rollout 3
this question is ambiguous. Nevertheless, likely answer \boxed{170} m/s. Alternatively,
Rollout 8
simplified.\n\nTherefore, the probability is 31/126, so m=31,
Top 16 Negative Activations
Rollout 1
able sequence, but assuming that S is any set. Hmm.\n\nAlternatively, since you can extend any or
Rollout 1
in S \right\} \) is an orthonormal system.\n\n1. **Orthonormal
Rollout 1
to see.\n\nAlternative idea. From the properties we earlier noted.\n\nIf we have to satisfy ⟨xi,
Rollout 1
This action would be the same as above for linear equations, but we already new that.\n\nGiven that,
Rollout 4
.003\n\nWait, no, actually. The term inside the sqrt for the first term is
Rollout 4
5000)^2= (4*10^6)/(25*10^
Rollout 6
but x varies and the bits accessed depend on x.\n\nAlternatively, perhaps encode the entire advice α_n into
Rollout 3
calculations would hold, but operations:\n\nmultiplying sf. If numbers are 8 (exact), R=
Rollout 1
iesz representation theorem cannot directly solve for the system unless... Wait, perhaps thinking of yi's dual variables.\n\n
Rollout 1
Let me consider the differences between the equations.\n\nFrom earlier, for x and x',\n\n||x - x
Rollout 3
15*66.512≈ let's compute exact:\n\n273 * 6
Rollout 1
perhaps given there exists a solution of the linear system described by ⟨x - y, x’ - y
Rollout 1
the only possible reason why the translations didn't work above is that the example I tried perhaps already assume a
Rollout 3
6.48*298= same approach: 66.48*30
Rollout 1
centroid.\n\nAlternatively, since in our deduction earlier, we found that if such a y exists for a set
Rollout 1
.\n\nWait, original problem: when translating back, orthonormal vectors must have that ⟨ x - y

DOWN_PROJ

Top 16 Positive Activations
Rollout 0
of such coprime pairs is \(2^k\), where \(k\) is the number of
Rollout 4
Then, take derivatives:\n\ndC_eq / C_eq = (dC1 / C1 + d
Rollout 2
coprime. Then, using the Chinese Remainder Theorem, combine the results. But before jumping
Rollout 2
9 again. 891*999 again: as before, 891*
Rollout 7
is twice the number of repeaters.\n\nTherefore, N = P + R = 2R + R
Rollout 2
is5 mod8. So according to Chinese Remainder Theorem, there is a unique solution mod1
Rollout 7
Since all participants are either promoted or repeaters, N = P + R.\n\nWe are given:\n\n1
Rollout 9
,0) to (10, 10). Then EM is the side from E (1
Rollout 8
Wait, for R^k G^(5 -k):\n\nNumber of ways to choose and arrange k red
Rollout 4
J. The maximum is ~15.1958 nJ, the minimum is ~1
Rollout 7
us an equation with x and N. Let's compute that.\n\nFirst expand:\n\n75*(2N
Rollout 3
slower speed, which aligns with ~168 m/s. So this checks out.\n\nSo all
Rollout 8
÷240=126. As above. Therefore 31/126.\n\n
Rollout 9
, 0). So the square is from (0,0) to (10, 1
Rollout 7
=> P= (2/3)N.\n\nYes that's correct. So that part is okay.\n\n
Rollout 3
Let's see, 168^2 = 28,224. 1
Top 16 Negative Activations
Rollout 5
12500 = 125y\n\nSimplify:\n\ny^2 - 2
Rollout 4
060,150 / 5025 �� let's divide 6,
Rollout 4
.000077 is 7.7e-5. sqrt(7.7
Rollout 2
10^1 - 1\n- 99 = 10^2 -
Rollout 6
given |x|, which is the length of x, then the combined machine would first compute α_{
Rollout 5
- 12500 = 125y\n\nSimplify:\n\ny^2 -
Rollout 2
1)...(10^{999} -1). We need to compute P mod
Rollout 3
,298. Then sqrt(225,298)= approx 474
Rollout 2
10^3=1000 mod125. 1000 divided by
Rollout 7
problem states that the pass mark was "fixed at 65". When all scores are increased by
Rollout 8
Wait, wait, no. Wait, the problem says "shuffles the 10 cards and lays
Rollout 1
y exists given that all pairwise distances in S are d.\n\nBut how can we construct such a y?\n\n
Rollout 5
8750 = 125x + 9375\n\nSubtract 1
Rollout 3
8,157.776 / 0.697 �� 26
Rollout 9
0 -100/h)*x ] from 0 to10.\n\nEvaluating at
Rollout 3
, could possibly round to 168 or 169. Maybe 168 m

Layer 43

GATE_PROJ

Top 16 Positive Activations
Rollout 9
Final Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(AIME\) has sides of length
Rollout 0
Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven a rational number between 0 and
Rollout 8
Answer**\n\boxed{157}\n\n<|im_start|>answer\nKathy has 5 red cards and
Rollout 3
boxed{169} m/s.\n\n<|im_start|>answer\nTo compute the mean molecular speed \( v \
Rollout 5
Answer**\n\boxed{181}\n\n<|im_start|>answer\nLet the shorter base be \( b \)
Rollout 4
{0.88\%}.\n\n<|im_start|>answer\nTo find the percentage error in the calculation of
Rollout 7
36}\n\n(b) No solution exists.\n\n<|im_start|>answer\nGiven a mathematics test with \( N \)
Rollout 2
Answer**\n\boxed{109}\n\n<|im_start|>answer\nTo find the remainder when \(9 \times
Rollout 1
} y \text{ exists as described}}\n\n<|im_start|>answer\nGiven an infinite-dimensional Hilbert space \( \
Rollout 0
, the number of rational numbers between 0 and 1 for which the product of the numerator and denominator
Rollout 4
_{\text{eq}} \) is:\n\n\[\n\frac{\Delta C_{\text{
Rollout 9
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides of length
Rollout 8
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red cards and
Rollout 7
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants is
Rollout 6
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $L \
Rollout 4
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values $C
Top 16 Negative Activations
Rollout 8
by dividing numerator and denominator by 240: 7440÷240=
Rollout 9
0] ( y_upper - y_lower ) dx\n\n= ��[0 to10] [
Rollout 3
8^2 = 28,224. 169^2 = 2
Rollout 8
, they each need to form a single block.\n\nSo if in the laid out sequence, there is at
Rollout 8
40÷240=126. As above. Therefore 31/12
Rollout 2
0^1 =10 mod125\n\n10^2=100 mod1
Rollout 8
0=1200\n\nk=2:\n\n2*P(5,2)*P(
Rollout 7
294/3=98 per student, but it's impossible since max original repeater is
Rollout 0
's either assigned entirely to a or to b.\n\nTherefore, 20! can be considered as a
Rollout 9
,0) to (10,10). That triangle would lie entirely inside the square?\n\nWait
Rollout 7
/6)=49*2=98.\n\nSame result, which is impossible. Therefore, the
Rollout 2
mod1000=109. So steps:\n\n9*99=891
Rollout 9
0, point G is at x= 0. For h=20, point G is at
Rollout 9
GEM would then be entirely within the square. Thus, the area in common would be the area of
Rollout 7
and <65. 98 not possible.\n\nN=24: x=2, sum
Rollout 9
h,5) which is inside the square. So triangle GEM would then be entirely within the square

UP_PROJ

Top 16 Positive Activations
Rollout 8
! ) / (5 - (5 -r))! ; wait, maybe it's better to think
Rollout 1
d / sqrt(2) times sqrt(2) from center y, which is exactly d. Wait
Rollout 1
of all these spheres (each centered at x ∈ S with radius d / sqrt(2)) is non
Rollout 1
simplex, in 3D, i.e., tetrahedron, but even a triangle is
Rollout 4
the 8 in the tenths place). Wait, percentage is per hundred, so 0.8
Rollout 1
0 for i ≠ j. But actually no, in the previous, if we take multiple points,
Rollout 8
5 * (5 *4 *3 *2 *1), but wait no:\n\nWait, for R
Rollout 2
^k-1 for anyk is always even? No. When k=1: 9 is
Rollout 9
Wait, EM is the top side of the square. Wait, hold on. \n\nWait, in the
Rollout 0
b have exponents for the primes? Wait no, in the problem statement, the numerator and denominator (
Rollout 8
all 5 red cards in order), but wait, she is laying out 5 cards out of
Rollout 4
%, which is half of 1.76% (max-min). But clearly, max-min range
Rollout 1
the intersection of all these spheres (each centered at x ∈ S with radius d / sqrt(2))
Rollout 1
center y, which is exactly d. Wait, no:\n\nWait, if e_x are orthonormal
Rollout 8
arranging all 5 red cards in order), but wait, she is laying out 5 cards out of
Rollout 4
0.003\n\nWait, no, actually. The term inside the sqrt for the first term
Top 16 Negative Activations
Rollout 7
5*(2N/3) = (75*2)/3 *N=150
Rollout 4
25.2004=603*25 +603*0.
Rollout 7
)=79*(16 +2)=79*18=1422.\n\nNon
Rollout 9
] dx\n\n= ��[0 to10] [ (1/5 x +8 +
Rollout 7
/3.\n\nSubstitute P and R:\n\n75*( (2N/3) + x )
Rollout 5
)/(75 + x)\n\nSo the equation becomes:\n\n(x + 175)(x -
Rollout 7
. Let's compute that.\n\nFirst expand:\n\n75*(2N/3) +75x
Rollout 7
denominator:\n\n62N/12 +54N/4 =62N/12
Rollout 7
- x.\n\nWe are told that after the increase:\n\n- The average of promoted becomes 75.
Rollout 7
=71N\n\nCalculate the terms:\n\n75*(2N/3) = (75
Rollout 3
73 �� 66.512 * 270 = 17,
Rollout 9
] dx\n\n= ��[0 to10] [ (2/5 x +6 )
Rollout 7
x =71N\n\nCalculate the terms:\n\n75*(2N/3) = (7
Rollout 3
293 = 66.512*(300 -7) = 6
Rollout 7
aters: R = N - P. Average 56.\n\nAfter score increase of 5:\n\n-
Rollout 4
=0.5 *1206e-12 *25.2004

DOWN_PROJ

Top 16 Positive Activations
Rollout 5
5H\n2. (x + 175)(H - h) = 12
Rollout 3
) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular speed
Rollout 3
in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular speed of
Rollout 9
10 is 100 -500/h.\n\nBut this is the overlapping area. Because
Rollout 5
5H\n2. (x + 175)(H - h) = 12
Rollout 9
, 10), M is at (10, 10), and E is at (
Rollout 5
-hand side:\n\ny^2 - 125y + 100y - 1
Rollout 9
] [5/h x +10 -50/h +5/h x -50/h]
Rollout 9
for an isosceles triangle with vertical base EM. Therefore, G is at (10 -
Rollout 3
16\n\nNow, 7,432.716 / 0.22
Rollout 4
combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage error
Rollout 9
-1000/h\n\n= 500/h +100 -100
Rollout 9
A is the bottom-left corner, I is the top-left, M is the top-right, and E
Rollout 9
2/2) + (10 -100/h)*x ] from 0 to1
Rollout 9
>10:\n\nWait 10 -50/h is less than10, because 5
Rollout 5
, area S = (75 + 175)/2 * H = 125
Top 16 Negative Activations
Rollout 7
they combine with x students whose original scores were between 60-64 (so with original average
Rollout 6
� P. Conversely, P �� P_bad-angel because we can ignore the angel string (set
Rollout 2
10^3=1000 mod125. 1000 divided by
Rollout 9
0 -100/h)*x ] from 0 to10.\n\nEvaluating at
Rollout 5
) = 125H\n\nFrom equation 1: h = (125H)/(
Rollout 7
problem states that the pass mark was "fixed at 65". When all scores are increased by
Rollout 9
10 - y)^2\n\nCanceling h^2:\n\ny^2 = (10 -
Rollout 9
10? Wait, if h is less than10, then G is at (10 -
Rollout 3
5 +15.87/(2*168.5)=15.87
Rollout 3
314*298/(π*0.028)) �� sqrt( (
Rollout 2
00.\n\nHence same asmodulo8 and125approach.\n\nTherefore, two separate methods
Rollout 3
8,157.776 / 0.697 �� 26
Rollout 3
, could possibly round to 168 or 169. Maybe 168 m
Rollout 8
it's P(5,3) * P(5,2) for the green and red cards
Rollout 2
uences. Therefore, the remainder upon division by 1000 is109. But
Rollout 6
angel. But according to the above, P_bad-angel = P, so NP = P_bad-

Layer 44

GATE_PROJ

Top 16 Positive Activations
Rollout 2
we consider the product modulo 8 and modulo 125 separately, then combine the results using the
Rollout 8
adjacent and all green cards laid out are adjacent.\n\nFirst, we calculate the total number of possible sequences of
Rollout 9
0)\), and \(E(10, 0)\).\n\n2. ** Isosceles
Rollout 0
factors of 20! are 2, 3, 5, 7, 1
Rollout 2
99 9's}}\) is divided by 1000, we consider the product modulo
Rollout 3
radon in kg/mol.\n\n**Steps:**\n\n1. **Molar Mass of Radon (R
Rollout 0
a \times b = 20!\).\n\n1. **Prime Factorization of 20!
Rollout 9
EM\) in \(\triangle GEM\).\n\n1. **Vertices of the Square**:\n -
Rollout 3
\times 8.314 \times 298 = 19,82
Rollout 3
\) of radon (Rn) gas, we use the formula for the mean (average)
Rollout 9
0, 0)\) to \(M(10, 10)\).\n - The
Rollout 3
mass of radon in kg/mol.\n\n**Steps:**\n\n1. **Molar Mass of Radon
Rollout 3
in Kelvin,\n- \( M \) is the molar mass of radon in kg/mol.\n\n**
Rollout 0
8}\n\n<|im_start|>answer\nGiven a rational number between 0 and 1, we need to write it
Rollout 8
answer\nKathy has 5 red cards and 5 green cards. She shuffles the 1
Rollout 1
} \) is an orthonormal system.\n\n1. **Orthonormal Conditions**:\n -
Top 16 Negative Activations
Rollout 8
/ (5 - (5 -r))! ; wait, maybe it's better to think:\n\nWait
Rollout 1
squared / 2 - ) ) / 2.\n\nBut not making progress.\n\nWait but if subtract the
Rollout 0
, then 5, 2, 1.\n\nWait, 18 for exponent of 2
Rollout 9
50/h), G's projection path, etc. Wait, perhaps getting complicated.\n\nAlternatively, maybe split
Rollout 4
+ 0.000064 = 0.0000769
Rollout 4
% and 0.004% error, but the voltage uncertainty was ±0.02
Rollout 7
answer for part(b) is N=24?\n\nWait, wait. Earlier we saw that in equations
Rollout 6
poly(2^{m}) = exponential in m. Therefore, the set S_L contains exponentially many strings
Rollout 6
p is a polynomial), which is exponential in m. Hence, the total number of strings of length m
Rollout 9
0). That triangle would lie entirely inside the square?\n\nWait no, if G is at (0,
Rollout 3
168.5≈ 168 or 169. In technical terms,
Rollout 0
*45=24^2*5).\n\nBut in general, n! will have prime factors
Rollout 6
2^{O(m)}, which is exponential in m. Therefore, S_L would not be sparse.\n\nTherefore
Rollout 5
18125/100 = 181.25, greatest integer
Rollout 1
y|| / sqrt{1 - a^, but this is stretching.\n\nBut since in each pair x
Rollout 9
- (something less than5)=more than5\n\nSo (0,10 -50/h

UP_PROJ

Top 16 Positive Activations
Rollout 3
) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular speed
Rollout 8
7200=7440. Therefore, 7440 /10*
Rollout 8
since the cards are of two colors, perhaps?\n\nWait, maybe not. If each card is unique (
Rollout 8
(10,5) but considering order.\n\nWait, but when they compute the probability, is the
Rollout 1
as the graph of a linear operator or something. Hmm.\n\nAnother approach. Assume H has an orthon
Rollout 2
\[\n x \equiv 5 \mod 8 \quad \text{and} \
Rollout 4
combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage error
Rollout 2
*(-1)= -3≡5 mod8.\n\nOkay, modulus8 seems fine. Good.\n\nNow modulus
Rollout 5
100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve
Rollout 0
a Unique Subsets of primes assigned to a.\n\nHence, since ordering for each subset, half will
Rollout 0
be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how many
Rollout 7
, original score >=60 would now pass.\n\nTherefore, in problem part(b), after increasing scores by
Rollout 2
000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here: I
Rollout 7
are original repeaters with original >=60. Thus, same as in part(a). So the contradiction
Rollout 1
||2 - ⟨x, t⟩ + C = 0.\n\nNot sure how helpful. If I
Rollout 1
.\n\nBut as we have already established from before.\n\nLet me reconsider the original distance condition. For any two
Top 16 Negative Activations
Rollout 7
is 56. Then, due to an error, all scores are increased by 5. After
Rollout 5
= 181.25, greatest integer not exceeding 181.25 is
Rollout 0
Let me check. The primes less than or equal to 20 are: 2, 3
Rollout 3
is the molar mass. Then there's the average speed, which is sqrt(8RT/(π
Rollout 9
(10 - y)^2]\n\nSince GE = GM, then:\n\nsqrt[h^2 + y
Rollout 9
to (10,10), the altitude to EM would just be the horizontal distance from G to
Rollout 3
8.5 m/s. Let's check with calculator steps.\n\nAlternatively, sqrt(28,4
Rollout 9
have an isosceles triangle GEM with base EM. So EM is one of the sides of
Rollout 5
181.25, so the greatest integer not exceeding that is 181.
Rollout 5
181.25, so the greatest integer not exceeding that is 181. Therefore
Rollout 9
So the answer is 25. Therefore box form \boxed{25}\n\nWait, that seems
Rollout 3
/s if we round 168.5 to two significant figures, but since temperatures are often given
Rollout 8
cross-verify.\n\nWait, let’s think of another approach. For instance, imagine that we consider the
Rollout 1
.\n\nNow, the problem wants me to find a point y such that when I take (sqrt(2
Rollout 1
d.\n\nNow, the problem wants me to find a point y such that when I take (sqrt(
Rollout 3
to temperature and molar mass. Wait, but the question specifically says "mean molecular speed"—is that

DOWN_PROJ

Top 16 Positive Activations
Rollout 6
if the advice is generated in polynomial time, then why not just incorporate that polynomial-time computation into the main
Rollout 9
confirm the positions. If the square is labeled AIME, moving around the square, then maybe A is
Rollout 7
itself isn't changed. Hmm. This is ambiguous.\n\nBut according to the problem statement: "due to
Rollout 1
find such a y by fitting it accordingly.\n\nSo why in the previous example with translated set S', it
Rollout 8
but since they are identical? Wait, wait, no. Wait, the problem says "shuffles the
Rollout 8
sequences. Let's consider two cases:\n\nCase 1: All 5 cards are red or all are
Rollout 0
but a and b have exponents for the primes? Wait no, in the problem statement, the numerator
Rollout 7
, but pass mark is fixed at 65. Wait, the scores are increased by 5,
Rollout 1
orthonormal systems can be uncountable. However, in that case, the problem still requires that
Rollout 2
for anyk is always even? No. When k=1: 9 is odd. 1
Rollout 7
increase. Hmm, so that's a critical point.\n\nWait, but reading the problem again: "due
Rollout 4
150 by 5025:\n\nFirst, 5025 * 12
Rollout 8
of such happy sequences. Let's consider two cases:\n\nCase 1: All 5 cards are red
Rollout 7
65?\n\nWait, this is a crucial point. Let's read the original problem again:\n\n"due
Rollout 3
. Let me confirm.\n\nRadon is a noble gas with atomic number 86. The atomic weight
Rollout 3
≈ 168 or 169. In technical terms, 168.5
Top 16 Negative Activations
Rollout 1
2 - 2c + d2)/2 = 0 ⇒ 0 = 0.\n\nThus
Rollout 0
}{b}\) and calculate the product of the resulting numerator and denominator. We need to determine how many
Rollout 9
2\n\nSubtract y^2 from both sides:\n\n0 = 100 - 20
Rollout 9
= ��[0 to10] [ (2/5 x +6 ) ] dx\n\n
Rollout 0
{a}{b}\) and calculate the product of the resulting numerator and denominator. We need to determine
Rollout 8
: 2!; total 2+2=4.\n\nCase2: for r=1,
Rollout 5
that is parallel to the bases and that divides the trapezoid into two regions of equal area.
Rollout 4
recalling how to calculate the equivalent capacitance of capacitors in series and then the energy stored. \n\nFirst
Rollout 1
2 = c - d2/2.\n\nSo substituting back into the previous equation:\n\n(2c
Rollout 5
b from both sides: b + 75 = 150; so b = 7
Rollout 1
⇒ (2c - d2)/2 - (c - d2/2) ) =
Rollout 0
written in lowest terms as a/b, the product of numerator and denominator is 20!=1×
Rollout 7
pass mark is 65. Originally, the average score of all participants is 66. The
Rollout 1
+ ||x'||2 - d2)/2 = 0.\n\nBut we also have that for any
Rollout 0
, 4 result in a < b and 4 in a >b. So when 72
Rollout 0
between 0 and 1 such that when written in lowest terms as a/b, the product of numerator

Layer 45

GATE_PROJ

Top 16 Positive Activations
Rollout 0
, corresponding to \boxed{128}? Or wait, wait prime factors of 20
Rollout 2
the remainder is109.\n\nTherefore, answer:109.\n\nBut let me recheck all
Rollout 4
approximately 0.88%.\n\nWait, maybe 0.88 is okay. Alternatively, let
Rollout 3
168 or 169. Maybe 168 m/s.\n\nBut checking further on
Rollout 5
is 181. Therefore, the answer is 181. \n\nWait, but let
Rollout 5
is 181. Therefore, final answer is 181. \n\nI think that's
Rollout 0
corresponding to \boxed{128}? Or wait, wait prime factors of 20!
Rollout 3
\boxed{170} m/s. Alternatively, could possibly round to 168 or
Rollout 4
0.88%, or maybe the answer is exact at 0.87%.\n\nBut wait,
Rollout 4
.8774964% ��0.8775%\n\nBut maybe doing
Rollout 3
\boxed{169} m/s. But need to specify.\n\nAlternatively, here's another angle
Rollout 3
68.547 m/s.\n\nSo approximately 168.5 m/s. Let
Rollout 3
Wait, but maybe the problem expects the answer as around 170 m/s. Alternatively, let
Rollout 9
\boxed{25}\n\nWait, that seems correct.\n\nLet me check this.\n\nGiven that the area
Rollout 0
So 8, so k=8. so answer is 2^(8-1)=12
Rollout 3
161.4 m/s. So approx 161 m/s.\n\nSo depending on
Top 16 Negative Activations
Rollout 0
<|im_start|>answer\nGiven a rational number between 0 and 1, we need to write it as a
Rollout 4
_2}{C_1 + C_2}\n\]\n\nGiven \( C_1 = 2
Rollout 6
$ .\n Note that a TM $M$ with oracle access to $S$
Rollout 2
9\cdots9}_{\text{999 9's}}\) is divided by
Rollout 6
_i \) is sparse, for each length \( n \), the number of strings in \( S \
Rollout 4
Delta C_{\text{eq}}}{C_{\text{eq}}} = \frac{4
Rollout 6
\) is sparse, for each length \( n \), the number of strings in \( S \)
Rollout 1
that the distance between any two distinct points in \( S \) is \( d \), we need to
Rollout 9
, 0)\) to \(M(10, 10)\).\n - The apex
Rollout 2
9 \times 99 \times 999 \times \cdots \times \under
Rollout 9
}{h}x + 10 - \frac{50}{h} \right) -
Rollout 4
= \frac{4.327}{1200} \approx 0.0
Rollout 8
or a combination of one block of red and one block of green.\n\n1. **Case 1:
Rollout 1
{H} \) and a set \( S \) of points such that the distance between any two
Rollout 9
to \(M(10, 10)\).\n - The apex \(G\) of the
Rollout 5
trapezoid properties. In a trapezoid, the midline (the segment connecting the

UP_PROJ

Top 16 Positive Activations
Rollout 7
"due to an error in the wording of a question, all scores are increased by 5. At
Rollout 9
square. Then the triangle GEM overlaps the square, and their common area is 80 square units
Rollout 7
: These are participants with original score <65, average 56. Their total score is
Rollout 7
"due to an error in the wording of a question, all scores are increased by 5. At
Rollout 7
at 65. Wait, the scores are increased by 5, so original scores become original +
Rollout 7
, 6 remaining. Original average of 2 students=62, so total score=62
Rollout 7
"due to an error in the wording of a question, all scores are increased by 5. At
Rollout 8
arranged in same color order, but different actual cards, it's still counts as the same color sequence.
Rollout 2
number of terms:999. The first two terms (k=1 andk=2)
Rollout 9
length we need to find.\n\nNow, the common area between triangle GEM and square AIME is
Rollout 7
due to an error in the wording of a question, all scores are increased by 5. At this
Rollout 8
in total. But when she lays out 5, those could be anywhere from 0 to 5
Rollout 7
. Wait, the scores are increased by 5, so original scores become original +5. Therefore,
Rollout 8
The key is when she lays them out in a row, orders matter, but whether different red cards are
Rollout 1
side length d. Suppose their positions are x1, x2, x3 forming a regular triangle in
Rollout 5
the height. \n\nSo, given that the two bases differ by 100 units, let me
Top 16 Negative Activations
Rollout 2
9\), so \(x \equiv 109 \mod 1000\).\n\n
Rollout 1
, but assuming that S is any set. Hmm.\n\nAlternatively, since you can extend any orthonormal
Rollout 3
=168^2 + 2*168*0.5 +0.2
Rollout 6
1^log n. But still, unsure.\n\nAlternatively, this is similar to problem 1.
Rollout 6
in S_L. However, this requires exponential time.\n\nAlternatively, assume the oracle can provide s_n in
Rollout 3
6.08=19,288.48 +199.53
Rollout 2
= 1000n + 109\), so \(x \equiv 1
Rollout 3
98 K (25°C) yields ~168.5 m/s. Therefore, since
Rollout 2
9 \times 99 \times 999 \times \cdots \times \under
Rollout 3
25,298)= approx 474.6 m/s. So about 4
Rollout 6
$S$ is bounded by $p(n)$ . In other words, $|S
Rollout 6
then query all possible extensions. But that's exponential.\n\nThis seems not helpful. Let's reorient.\n\n
Rollout 3
6=18,157.776\n\nPlus 0.15*6
Rollout 1
balls are not compact hence there is no guarantee.\n\nHences perhaps requires another approach, using the structure of
Rollout 3
. Rounded to 169 m/s.\n\nHence, some approximations.\n\nBut given all possible
Rollout 1
However, in infinitely many dimensions, such an argument does not directly apply.\n\nAlternatively, given that for each

DOWN_PROJ

Top 16 Positive Activations
Rollout 8
9 * 8 * 7 * 6. Wait, because each card is drawn without replacement
Rollout 3
there's a formula for the root mean square speed, which is related to temperature and molar mass.
Rollout 8
of ways to assign the red cards to the red positions and green cards to the green positions. Let me
Rollout 7
original groups and the new groups.\n\nLet me denote variables:\n\nOriginal scenario:\n\n- Number of original promoted:
Rollout 5
trapezoid into areas with ratio 2:3, leading us to bases of 75
Rollout 5
0) + (b + 100))/2 * (h/2) = (2
Rollout 7
+ x.\n\nSimilarly, new repeater count = R - x.\n\nWe are told that after the increase
Rollout 7
2)= (14/3)*(12/N)*(N)=14/3*12
Rollout 8
6. So 31/126.\n\nWait, wait: 7440 �
Rollout 1
system must be countable because the space is separable? Wait, but the problem states that S is
Rollout 6
possible candidates for the suffix, as it is exponential in p(n).\n\nAlternatively, since the original M in
Rollout 7
2. Total promoted-from-repeaters=448 -252=196. So
Rollout 1
must be countable because the space is separable? Wait, but the problem states that S is not
Rollout 6
the advice is fixed for all inputs of length n, and the machine runs in poly-time. But in
Rollout 8
5 red cards, but since they are identical? Wait, wait, no. Wait, the problem
Rollout 0
5? 16 <45: yes.\n\n3. Assign 3 to a, others to
Top 16 Negative Activations
Rollout 1
is d, I need to find a point y such that { sqrt(2)/d (x -
Rollout 6
this case, the oracle TM can, given input x of length n, retrieve s_n from the oracle
Rollout 3
some books, they use 273 K, others 298 K. For example,
Rollout 4
0 -60,000 +150=5,940,15
Rollout 8
color. Otherwise, if the cards were unique, it would matter which specific red or green card is where
Rollout 1
H all pairwise distance d apart. To find y such that ��2/d (x - y)
Rollout 1
y|| = d / sqrt(2),\n\nwe also have ||x - x'|| = d = sqrt
Rollout 1
∈S} is orthonormal.\n\nWhich is equivalent to requiring that { x - y } is orthogonal
Rollout 0
1 and a*b=20!. Hence, equivalent to coprime pairs (a,b) with
Rollout 4
ature with coefficients. Then the total energy error is sqrt(0.362 + (2*
Rollout 6
are represented in binary, then for each n and i < p(n), the length of the string would
Rollout 4
.0036)^2 = 0.00001296\n\nAdding
Rollout 4
5)/(2010 + 3015). Let's compute that:\n\n201
Rollout 1
i S with mutual distance d, can you find such a y presumably by retro-fitting the construction.\n\n
Rollout 3
. In chemistry, STP is 0°C (273.15 K) and
Rollout 8
are of two colors, perhaps?\n\nWait, maybe not. If each card is unique (distinct), even

Layer 46

GATE_PROJ

Top 16 Positive Activations
Rollout 9
Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(AIME\) has sides of length
Rollout 3
{169} m/s.\n\n<|im_start|>answer\nTo compute the mean molecular speed \( v \)
Rollout 1
} y \text{ exists as described}}\n\n<|im_start|>answer\nGiven an infinite-dimensional Hilbert space \( \
Rollout 4
0.88\%}.\n\n<|im_start|>answer\nTo find the percentage error in the calculation of the
Rollout 2
**\n\boxed{109}\n\n<|im_start|>answer\nTo find the remainder when \(9 \times
Rollout 0
**\n\boxed{128}\n\n<|im_start|>answer\nGiven a rational number between 0 and 1
Rollout 1
y \text{ exists as described}}\n\n<|im_start|>answer\nGiven an infinite-dimensional Hilbert space \( \math
Rollout 5
**\n\boxed{181}\n\n<|im_start|>answer\nLet the shorter base be \( b \) and
Rollout 5
\boxed{181}\n\n<|im_start|>answer\nLet the shorter base be \( b \) and the
Rollout 8
**\n\boxed{157}\n\n<|im_start|>answer\nKathy has 5 red cards and 5
Rollout 9
Final Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(AIME\) has sides of length
Rollout 3
boxed{169} m/s.\n\n<|im_start|>answer\nTo compute the mean molecular speed \( v \
Rollout 5
Answer**\n\boxed{181}\n\n<|im_start|>answer\nLet the shorter base be \( b \)
Rollout 8
Answer**\n\boxed{157}\n\n<|im_start|>answer\nKathy has 5 red cards and
Rollout 0
Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven a rational number between 0 and
Rollout 4
{0.88\%}.\n\n<|im_start|>answer\nTo find the percentage error in the calculation of
Top 16 Negative Activations
Rollout 2
P ��109 mod125.\n\nSo summary so far:\n\nP≡5 mod8
Rollout 2
k -1) �� -1 mod8.\n\nTherefore, for k=1: (2^
Rollout 0
,17,19 have exponent 1.\n\nSo primes 2: 18,
Rollout 2
)^{997} mod125\n\nBecause from k=3 to k=99
Rollout 3
^2 = 28,561. So sqrt(28,436)
Rollout 8
either order), or all reds or all greens.\n\nSo first, we can calculate the number of such
Rollout 0
= 0), so that's exponent 2.\n\nWait, 7^2 divides into 2
Rollout 8
acceptable, like RRRRR or GGGGG.\n\nSo in total, the happy sequences are those that
Rollout 4
%. So approximately 0.8776%.\n\nThus, the percentage error is approximately 0.
Rollout 4
percentage: 0.87755%. So approximately 0.8776%.\n\n
Rollout 8
440 / 30240.\n\nSimplify this fraction.\n\nFirst, divide numerator and
Rollout 8
layout, they each need to form a single block.\n\nSo if in the laid out sequence, there is
Rollout 4
but multiplied by 2 gives 0.8%.\n\nSo, when combining these: sqrt( (0
Rollout 2
and P≡109 mod125.\n\nNow need to find a number x such that:\n\n
Rollout 2
k for k ≥3 is 0 mod8. Thus, for k ≥3, 2^
Rollout 2
-1 �� -1 mod125.\n\nTherefore, product P mod125 is:

UP_PROJ

Top 16 Positive Activations
Rollout 4
0 + 3000} = \frac{6000000}{5
Rollout 4
000 + 3000} = \frac{6000000
Rollout 2
000\n \]\n\nThus, the remainder when the product is divided by 100
Rollout 9
ies h = 25\n \]\n\nThus, the length of the altitude to \(EM\)
Rollout 0
= 128\).\n\nThus, the number of rational numbers between 0 and 1 for
Rollout 2
mod 1000\n \]\n\nThus, the remainder when the product is divided by
Rollout 5
625}{2}} = \sqrt{\frac{36250}{2}} =
Rollout 9
25\n \]\n\nThus, the length of the altitude to \(EM\) in \(\
Rollout 5
2(b + 75) \implies 3b + 75 = 2b
Rollout 9
to \(EM\) in \(\triangle GEM\) is \(\boxed{25}\).<|im_end|>
Rollout 2
\times 3 \times (-1)^{997} \equiv 1 \times
Rollout 4
2000 + 3000} = \frac{600000
Rollout 8
126 = 157\).\n\n\[\n\boxed{157}\n\]
Rollout 2
1000\n \]\n\nThus, the remainder when the product is divided by 10
Rollout 5
5) = 2(b + 75) \implies 3b + 75
Rollout 4
\frac{6000000}{5000} = 120
Top 16 Negative Activations
Rollout 0
, 3^8, 5^4, etc., up to the primes less than or equal
Rollout 3
the molar mass would be approximately 222 g/mol, right?\n\nSo, first, M
Rollout 3
the gas constant, which is 8.314 J/(mol·K).\n\nAssuming standard
Rollout 3
6, so atomic weight is approximately 222 (since the most stable isotope is radon
Rollout 8
- For each possible number of red cards (0 <= r <=5):\n\nIf r=0: all
Rollout 3
average speed is sqrt(8RT/(πM)).\n\nBut if the problem doesn't give a temperature
Rollout 5
of equal area, the length x can be found using the formula which is sort of a weighted average.
Rollout 7
, this implies N must be divisible by 12 (so that x is an integer). Therefore,
Rollout 2
10^k is 0 mod125?\n\nWait, 10^1 =1
Rollout 3
gas constant, which is 8.314 J/(mol·K).\n\nAssuming standard temperature
Rollout 8
sequences. Let's consider two cases:\n\nCase 1: All 5 cards are red or all are
Rollout 3
, which is indeed sqrt(8RT/(πM)). The root mean square speed is higher, as
Rollout 3
?\n\nSo, first, M = 222 g/mol = 0.222 kg
Rollout 5
is the average of the squares of the two bases. Wait, no, let me check. \n\nI
Rollout 3
is indeed given by sqrt(8RT/(πM)), whereas the root mean square speed is sqrt(
Rollout 3
indeed given by sqrt(8RT/(πM)), whereas the root mean square speed is sqrt(3

DOWN_PROJ

Top 16 Positive Activations
Rollout 4
Let me calculate 4.327 divided by 1200.\n\nDivide by
Rollout 4
compute that more accurately.\n\n4.327 / 1200. Let me calculate
Rollout 3
=66.48*298= same approach: 66.48*
Rollout 3
6.512 * 298:\n\nLet me compute this multiplication.\n\n66.5
Rollout 4
0 / 5025. \n\nDividing 6,060,150
Rollout 4
1990 * 2985= (2000 -10)*(3
Rollout 3
66.512*293=66.512*(200
Rollout 3
2 * 298 = let's compute 66.512 * 30
Rollout 4
7 divided by 1200.\n\nDivide by 100: 4.3
Rollout 2
9= 891*999= let's compute: 891*(1
Rollout 3
6.512 * 293 = 66.512*(30
Rollout 3
.512 * 293 = 66.512*(300
Rollout 4
,150 by 5025:\n\nFirst, 5025 * 1
Rollout 4
. Let me calculate 4.327 divided by 1200.\n\nDivide by
Rollout 3
66.512*273=66.512*(200
Rollout 3
.512 *293= compute 66.512*290
Top 16 Negative Activations
Rollout 7
, x=N/12.\n\nTherefore:\n\nA*(N/12)=14N/3
Rollout 7
2, x=1, sum=98*1=98. While student’s original score
Rollout 0
(20/27)=6 +2 +0=8\n\nFor prime 5: floor
Rollout 7
=12.\n\nx=36/12=3.\n\nOriginal repeaters:12 students
Rollout 7
R=8.\n\nx=24/12=2.\n\nOriginal repeaters:8 students.
Rollout 7
so total score=62*2=124. Remaining 6 students average 54
Rollout 7
aters: 62*1 +54*3=62 +162=2
Rollout 7
original score >=60 (since 60 +5 =65) are promoted. Therefore,
Rollout 1
each of length d / sqrt(} and whose differences have length d.\n\nBecause such vectors could form an
Rollout 9
6x ] from0 to10\n\n= (1/5)(100) +6
Rollout 9
0 to10\n\n= (1/5)(100) +60 -0 =
Rollout 9
(10,0).\n\nDistance GE = sqrt[(10 - (10 - h))^2
Rollout 8
sequences = 2 *5! + Sum_{r=1}^4 [2 * P(
Rollout 1
x0, we get z0 = x0 - x0 = 0.\n\nDoes that help?\n\n
Rollout 9
x ] from0 to10\n\n= (1/5)(100) +60
Rollout 7
1 +54*3=62 +162=224. Which is

Layer 47

GATE_PROJ

Top 16 Positive Activations
Rollout 1
\mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that
Rollout 8
Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need
Rollout 3
Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean
Rollout 6
L$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through
Rollout 4
in this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the
Rollout 0
!$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out
Rollout 9
in $\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this
Rollout 5
2/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need
Rollout 2
$1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here
Rollout 7
oted $47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve
Rollout 2
Final Answer**\n\boxed{109}\n\n<|im_start|>answer\nTo find the remainder when \(9 \
Rollout 4
2}\n\]\n\nSubstituting the values:\n\n\[\n\text{Relative error in } E =
Rollout 1
point } y \text{ exists as described}}\n\n<|im_start|>answer\nGiven an infinite-dimensional Hilbert space \(
Rollout 9
**Final Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(AIME\) has sides of
Rollout 8
Final Answer**\n\boxed{157}\n\n<|im_start|>answer\nKathy has 5 red cards and
Rollout 3
{169} m/s.\n\n<|im_start|>answer\nTo compute the mean molecular speed \( v \)
Top 16 Negative Activations
Rollout 8
126. Then check if they can be simplified. Since 31 is prime. 1
Rollout 3
,820.576.\n\nDivide that by π * 0.222
Rollout 3
). Let's see, 168^2 = 28,224.
Rollout 8
��240= 126. So 31/126.\n\nWait
Rollout 3
28,436. Then take square root.\n\nsqrt(28,436).
Rollout 8
check by dividing numerator and denominator by 240: 7440÷240
Rollout 8
26. Then check if they can be simplified. Since 31 is prime. 12
Rollout 3
22). Wait, Radon has several isotopes, but the most common one in decay chains is
Rollout 8
Therefore, 31/126. Then check if they can be simplified. Since 3
Rollout 0
of them, 4 pairs would have a < b. Let's check. Possible assignments:\n\n1.
Rollout 0
20! but 7^3 does not.\n\nBut exponent of 7 is 2,
Rollout 0
but how many of those assignments result in a < b?\n\nSo for 720, with prime
Rollout 8
7440\n\nHence, the probability is 7440 / 302
Rollout 2
0,109. mod1000, that's109. So 8
Rollout 0
divides into 20! but 7^3 does not.\n\nBut exponent of 7 is
Rollout 2
9.\n\n9*99=891. 891 mod125:

UP_PROJ

Top 16 Positive Activations
Rollout 0
4,7:2, which are all even? Wait, 18,8,4,
Rollout 7
8, but >=60 and <65. 98 not possible.\n\nN=24
Rollout 4
000^2) )^2 )\n\nWait, this is another way of writing the terms.
Rollout 7
=65 -5 to64 -5(wait no). wait: new score >=65.
Rollout 9
/h)x + (50/h) +0? Wait, let's verify:\n\nFrom G(1
Rollout 7
score + 5 must be >=65? Wait, but if a student's original score +5
Rollout 2
=999*10 +9, wait, perhaps 9999=9*
Rollout 7
(56 -54)N/3 ?\n\nWait, wait:\n\nWait 18N is
Rollout 4
52 +0.52 +something)%? Wait, no—wait:\n\nWait, the C
Rollout 3
≈ 168 or 169. In technical terms, 168.5
Rollout 0
2,3,5,7 are even?\n\nWait floor operations:\n\nFor prime 2:
Rollout 7
12:\n\nA -54 =8.\n\nTherefore,A =62.\n\nSo the average original
Rollout 4
is [0.003]^2. But wait:\n\nFirst term inside the formula is [ (
Rollout 9
)x + (50/h) +0? Wait, let's verify:\n\nFrom G(10
Rollout 7
70, or is it still 65?\n\nWait, this is a crucial point. Let's
Rollout 8
(5 - (5 -r))! ; wait, maybe it's better to think:\n\nWait,
Top 16 Negative Activations
Rollout 0
}\) and calculate the product of the resulting numerator and denominator. We need to determine how many such rational
Rollout 1
\in \mathcal{H} \) such that the set \( \left\{ \frac
Rollout 6
machine \( M \). By the Meyer's theorem equivalence between \( \textbf{P}/\text
Rollout 3
oltzmann distribution:\n\n\[\nv = \sqrt{\frac{8RT}{\pi M}}\n
Rollout 6
i\rangle\) and checks membership in \( S \).\n\n---\n\n**Problem 2:**\n\nFor \
Rollout 1
in \mathcal{H} \) such that the set \( \left\{ \frac{\
Rollout 1
d, the resulting set is an orthonormal system. That means each vector has length 1 and
Rollout 6
this string, and accepts if the oracle confirms its presence.\n\n**Answer:** \nThe sparse set \( S
Rollout 1
right\} \) is an orthonormal system.\n\n1. **Orthonormal Conditions**:\n
Rollout 8
them in a row. We need to find the probability that all red cards laid out are adjacent and all
Rollout 5
to find x, which is the length of the segment that divides the trapezoid into two equal
Rollout 2
00 is \(\boxed{109}\).<|im_end|>
Rollout 5
divides the trapezoid into two regions of equal area, we use the root mean square of the
Rollout 1
and a set \( S \) of points such that the distance between any two distinct points in \( S
Rollout 3
169} m/s.\n\n<|im_start|>answer\nTo compute the mean molecular speed \( v \) of
Rollout 1
distance between any two distinct points in \( S \) is \( d \), we need to show there

DOWN_PROJ

Top 16 Positive Activations
Rollout 4
0.52 +0.52 +something)%? Wait, no—wait:\n\nWait,
Rollout 4
pF. That is 4.326/1200≈0.00
Rollout 7
=98 is impossible.\n\nTherefore, inconsistency. What does that mean? It means that our assumptions might
Rollout 4
C_eq was 4.327 pF, C_eq is 1200 p
Rollout 2
mod8.\n\nTherefore, equation becomes:\n\n(5m +5)≡5 mod8\n\nSubtract
Rollout 7
score + 5 must be >=65? Wait, but if a student's original score +5
Rollout 3
)^2=168^2 + 2*168*0.5 +0
Rollout 1
for all x, x' in S?\n\nSuppose that all points in S lie on a sphere in
Rollout 0
4,7:2, which are all even? Wait, 18,8,4,
Rollout 9
80, then h would be 16. But 10 max, which is inconsistent.
Rollout 4
.2004=603*25 +603*0.20
Rollout 5
Let me denote the original trapezoid with bases a = 75, b = 1
Rollout 0
6. However, because 20! is not a square (due to the primes with exponents
Rollout 1
by defining various terms. Let me see.\n\nSuppose that all points in S lie in the set {
Rollout 3
273 with R=8.314:\n\nWe did 161 m/s.\n\n
Rollout 4
uncertainty of ±6 pF. This is different. Which one is correct?\n\nWait, but when dealing
Top 16 Negative Activations
Rollout 0
2,3,5:\n\nThe 8 assignments:\n\n1. Assign none to a => a=1
Rollout 3
, I need to recall. In some problems, if the temperature isn't given, it's assumed to
Rollout 4
00 ± 0.02 V. Hmm, alright, let's start by recalling how to
Rollout 6
I'm going in circles here. Given the time constraints, maybe for problem 3, since it's
Rollout 4
5) is sqrt(10^-5) = 10^-2.5 ��
Rollout 2
\n\n2^3 = 0 mod 8\n\n2^4 = 0 mod 8\n\n
Rollout 6
some advice string α_n.\n\n\nI think I'm going in circles here. Given the time constraints, maybe
Rollout 5
points of the legs divides the trapezoid into two regions with areas in the ratio 2:
Rollout 1
when I take each x in S, subtract y, scale by sqrt(2)/d, the resulting
Rollout 6
$n \in \mathbb{N}$ , the number of strings of length $n$
Rollout 2
25. The number of such terms is999-2=997. Hence,
Rollout 3
calculate the mean molecular speed of radon gas. Hmm, first, I remember there's a formula for
Rollout 9
first. \n\nLet me draw the square AIME. Since it's a square, all sides are
Rollout 8
red and green counts (from 1 to 4 reds and 4 to 1 greens).\n\n
Rollout 9
10 -100/h)*10\n\n= (10/h)*(50) +
Rollout 6
would be O(log n + log p(n)) = O(log n) (since p(n) is

Layer 48

GATE_PROJ

Top 16 Positive Activations
Rollout 9
Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(AIME\) has sides of length
Rollout 0
**\n\boxed{128}\n\n<|im_start|>answer\nGiven a rational number between 0 and 1
Rollout 1
y \text{ exists as described}}\n\n<|im_start|>answer\nGiven an infinite-dimensional Hilbert space \( \math
Rollout 2
**\n\boxed{109}\n\n<|im_start|>answer\nTo find the remainder when \(9 \times
Rollout 4
0.88\%}.\n\n<|im_start|>answer\nTo find the percentage error in the calculation of the
Rollout 8
**\n\boxed{157}\n\n<|im_start|>answer\nKathy has 5 red cards and 5
Rollout 5
**\n\boxed{181}\n\n<|im_start|>answer\nLet the shorter base be \( b \) and
Rollout 3
{169} m/s.\n\n<|im_start|>answer\nTo compute the mean molecular speed \( v \)
Rollout 2
Answer**\n\boxed{109}\n\n<|im_start|>answer\nTo find the remainder when \(9 \times
Rollout 1
} y \text{ exists as described}}\n\n<|im_start|>answer\nGiven an infinite-dimensional Hilbert space \( \
Rollout 7
6}\n\n(b) No solution exists.\n\n<|im_start|>answer\nGiven a mathematics test with \( N \) participants
Rollout 9
\nSquare \(AIME\) has sides of length 10 units. Isosceles triangle \(
Rollout 0
Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven a rational number between 0 and
Rollout 5
100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve
Rollout 3
on (Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate
Rollout 1
$ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show
Top 16 Negative Activations
Rollout 4
49e-3*sqrt(10). Because sqrt(10)=3.16
Rollout 4
Wait, this is another way of writing the terms. Maybe this will increase precision?\n\nWait, or rather
Rollout 1
property) and maybe in a weakly compact set. But in Hilbert spaces, closed balls are weak
Rollout 6
to use the same number of bits for each n. Wait, but log n is variable length. If
Rollout 1
Hilbert spaces, closed balls are weakly compact.\n\nBut given that spheres are closed and convex. Wait
Rollout 4
000^2) )^2 )\n\nWait, this is another way of writing the terms.
Rollout 5
00 = 125y\n\nSimplify:\n\ny^2 - 25y -
Rollout 1
)/2.\n\nTherefore, substituting into the previous equation:\n\n⟨x - x', y⟩ = (
Rollout 9
height.\n\nBut the bases are the two vertical sides. Wait, no, here the 'height' in
Rollout 9
, which is inconsistent. So that suggests something's wrong.\n\nWait, perhaps my assumption is incorrect. Let
Rollout 1
this as the graph of a linear operator or something. Hmm.\n\nAnother approach. Assume H has an or
Rollout 3
31 instead of 8.314. Let me check. Suppose R=8.3
Rollout 6
) bits, but pad it to a fixed length. But perhaps in the original problem, there is no
Rollout 4
lead to higher.\n\nWait, compute both ways:\n\nLet's compute the maximum energy. To get maximum energy
Rollout 9
is 10; the bases are parallel sides? But wait the two lines at x=0 and
Rollout 6
use the same number of bits for each n. Wait, but log n is variable length. If instead

UP_PROJ

Top 16 Positive Activations
Rollout 0
a share no primes with b. Therefore, the coprime pairs (a,b) with a*b
Rollout 0
, both positive integers, a < b.\n\nAnd this number is equal to 2^{k -1
Rollout 5
= 181.25, greatest integer not exceeding 181.25 is
Rollout 5
181.25, greatest integer not exceeding 181.25 is 1
Rollout 3
/(πM)). The problem mentions "mean molecular speed," which is usually the average speed. Wait,
Rollout 5
5 is 181. Therefore, final answer is 181. \n\nI think that
Rollout 0
20! and a < b.\n\nThus previous logic applies; each such pair is obtained by assignment of
Rollout 0
\(a\) and \(b\) are coprime factors of 20!. So they form a
Rollout 0
=N!:\n\nSince a and b are coprime.\n\nHence equivalence between choices over subsets of primes
Rollout 0
\(a = b\). So all coprime pairs with \(a \times b = 2
Rollout 0
when we write a and b as coprime factors, the way I'm thinking is that the assignment
Rollout 0
When we say that a and b are coprime with a*b=20!, then a and
Rollout 5
181.25, so the greatest integer not exceeding that is 181. Therefore
Rollout 5
1.25, so the greatest integer not exceeding that is 181. Therefore, the
Rollout 9
10x10, the area of overlap between triangle GEM and square AIME is
Rollout 8
perhaps not). The key is when she lays them out in a row, orders matter, but whether different
Top 16 Negative Activations
Rollout 7
= (56 -54)N/3 ?\n\nWait, wait:\n\nWait 18N
Rollout 7
P =79x - (A +5)x\n\n-3P=(74 -A)
Rollout 2
2^k \equiv 0 \mod 8\), so \(2^k - 1
Rollout 4
}{(C_1 + C_2)^2} \quad \text{and} \quad
Rollout 7
213N -209N)/3=4N/3.\n\nTherefore:\n\n16
Rollout 7
scores=65 -5 to64 -5(wait no). wait: new score >=65
Rollout 5
12500 = 125y\n\nBring all terms to left side:\n\ny^
Rollout 4
(C_1 + C_2)^2} \quad \text{and} \quad \
Rollout 7
+A x=79P +79x\n\n→-3P +A x -7
Rollout 7
213N -205N)/3=8N/3.\n\nHence:\n\n3
Rollout 5
12500 = 125y\n\nSimplify:\n\ny^2 - 2
Rollout 2
^k \equiv 0 \mod 8\), so \(2^k - 1 \
Rollout 2
) \times (-1)^{997} \equiv 1 \times 3 \times
Rollout 1
+ (d2 / 2)) / 2 = (3d2 / 2) /
Rollout 5
5) = 2(b + 75) \implies 3b + 75
Rollout 7
/2= (28 +21)/6=49/6≈8.16

DOWN_PROJ

Top 16 Positive Activations
Rollout 2
$1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here
Rollout 8
Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need
Rollout 7
oted $47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve
Rollout 5
2/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need
Rollout 9
in $\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this
Rollout 6
L$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through
Rollout 4
in this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the
Rollout 0
!$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out
Rollout 3
Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean
Rollout 1
\mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that
Rollout 3
temperature is given, perhaps they just want the formula, but I don't think so. Wait, if
Rollout 6
) to decide \( x \in L \).<|im_end|>
Rollout 3
for multiplication/division, the number with the least sf determines, which is 3. So answer should
Rollout 3
, can't access it.\n\nAlternatively, in absence of the temperature, perhaps mention the issue, but given
Rollout 7
{36}\n\n(b) No solution exists.\n\n<|im_start|>answer\nGiven a mathematics test with \( N \
Rollout 7
(A +5)x\n\n-3P=(74 -A) x.\n\nWait, substitute P
Top 16 Negative Activations
Rollout 2
9= 891*999= let's compute: 891*(1
Rollout 2
9: 891*999=891*(1000 -1
Rollout 3
6.512 * 298 = ?\n\nAs before:\n\n66.512
Rollout 4
5). Well, sqrt(7.7) is approx 2.7749, so
Rollout 7
5 -59)x=16x.\n\nTherefore, the equation becomes:\n\n209N/
Rollout 0
so that's included.\n\nBut if a and b can be written as a = 1 and b =
Rollout 8
she has 5 red cards and 5 green cards, with each color being indistinct, then
Rollout 7
75 -59)x=16x.\n\nTherefore, the equation becomes:\n\n209N
Rollout 3
66.512*293=66.512*(200
Rollout 1
, same for ||x' - y||2. Therefore:\n\nd2 = (d2)/2
Rollout 8
P(5,3)=2*(5*4)*(5*4*3)=2*2
Rollout 6
think about P_angel: the original TM M is using α_n to decide L, so for an
Rollout 3
66.512 *293= compute 66.512*2
Rollout 3
*8.314*298)/(3.1416*0.0
Rollout 3
8 * R * T) / (π * M))\n\nR = 8.314 J
Rollout 3
66.512*273=66.512*(200

Layer 49

GATE_PROJ

Top 16 Positive Activations
Rollout 6
then O(n^k / log n), but this is still not polynomial per length. Wait.\n\nAlternatively
Rollout 6
have p(n) strings for each n, but this might not be sparse if p(n) is a
Rollout 2
that 109 mod8=5, which satisfies the first congruence.\n\nWait, hmm
Rollout 2
109 mod8 is 5, which matches. So 109 is indeed congr
Rollout 6
their values. Wait, that was the prior idea which leads to non-sparsity.\n\nAlternatively, define S
Rollout 6
an index (n, i). But again, this requires handling indices, and similar length as before.\n\n
Rollout 2
: 109 mod8=5, which matches. Therefore, the solution is x≡1
Rollout 6
(n,i) with their values. Wait, that was the prior idea which leads to non-sparsity
Rollout 6
strings of length m is at most 1, which is sparse. Then, the set S_L =
Rollout 0
this would not satisfy a < b, therefore this assignment would not be included in the count. So how
Rollout 0
, 2^(3 -1)=4, which matches what we saw for 720.\n\n
Rollout 6
n and position, hence for each n, the number of strings is O(p(n)/log n),
Rollout 8
4 (since when k=0, all green, and k=5, all red, which are
Rollout 6
(m)}, which is not polynomial.\n\nTherefore, this encoding causes S_L to be not sparse. Alternatively,
Rollout 0
.\n\nBut exponent of 7 is 2, which is even. Wait a second. Then primes
Rollout 8
would occur only if both cards are red: The number of such sequences is 2*1 =2
Top 16 Negative Activations
Rollout 3
(Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the
Rollout 4
stored in this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find
Rollout 3
) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular speed
Rollout 3
on (Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate
Rollout 1
\mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that
Rollout 3
Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean
Rollout 4
energy stored in this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to
Rollout 1
$ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show
Rollout 6
$L$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to
Rollout 2
$1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here
Rollout 4
in this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the
Rollout 5
100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve
Rollout 7
oted $47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve
Rollout 8
Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need
Rollout 9
in $\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this
Rollout 4
combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage error

UP_PROJ

Top 16 Positive Activations
Rollout 6
S is sparse.\n\nThen, the oracle machine M, when given <x,i>, encodes i and
Rollout 6
bit of α_n. Then, the machine M, given x of length n, needs to get the
Rollout 6
sparse.\n\nTo retrieve α_n, the oracle machine, given input x of length n, would need to
Rollout 0
how many rational numbers between 0 and 1", each such number is associated with a unique pair (
Rollout 6
S_i, then S is sparse. Then, when given <x, i>, the machine encodes
Rollout 1
0.\n\nCalculating the norm of v_x: ||v_x|| = (sqrt(2)/
Rollout 6
Thus, S is sparse.\n\nThen, the oracle machine M, when given <x,i>, encodes
Rollout 6
in S_i, then S is sparse. Then, when given <x, i>, the machine enc
Rollout 6
is sparse.\n\nThen, the oracle machine M, when given <x,i>, encodes i and x
Rollout 3
25°C for average speeds.\n\nHence, if following such an example, this question's answer would
Rollout 0
\(a \times b = 20!\), then each of the prime powers in the factorization
Rollout 6
, S is sparse.\n\nThen, the oracle machine M, when given <x,i>, encodes i
Rollout 6
Then, when given <x, i>, the machine encodes (i,x) into a string s
Rollout 9
Wait no, if the altitude is from G to EM, which is length h, then if h is
Rollout 8
's better to think:\n\nWait, arranging r reds and 5−r greens in order where all
Rollout 6
k.\n\nSo yes, in this case, if we construct S as the union of tagged strings (i
Top 16 Negative Activations
Rollout 4
} = \sqrt{12.96 + 5.76} = \sqrt{
Rollout 5
3b + 75 = 2b + 150 \implies b =
Rollout 7
=486. Total=186+486=672=56
Rollout 3
so 19,953.6 - 465.584 =1
Rollout 3
pi*0.028= ~0.08796\n\n19,8
Rollout 7
05N/3=(213N -205N)/3=8N/
Rollout 4
42)=sqrt(12.96+5.76)=sqrt(18.
Rollout 3
*0.028= ~0.08796\n\n19,82
Rollout 4
42)=sqrt(12.96+5.76)=sqrt(18.
Rollout 3
, so ~7,482.6 - 49.884 = 7
Rollout 5
3b + 75 = 2b + 150 \implies b =
Rollout 8
5*4*3)=2*20*60=2400\n\nk=
Rollout 2
solution is x≡109 mod1000.\n\nWait, but hold on, earlier we
Rollout 4
00013 + 0.000064) �� sqrt(0
Rollout 5
3b + 75 = 2b + 150; subtract 2b
Rollout 8
(5,3))=2*20*60=2400.\n\nSame for

DOWN_PROJ

Top 16 Positive Activations
Rollout 4
3*25.2004=603*25 +603*
Rollout 4
004e-12 �� 603*25.2004
Rollout 8
denominator by 240: numerator becomes 7440 �� 240=
Rollout 5
zoid is $100$ units longer than the other base. The segment that joins the mid
Rollout 0
fraction \(\frac{a}{b}\) where \(0 < a < b\) and \(a
Rollout 6
. In this case, since the angel string is fixed for each input length, and computed deterministically,
Rollout 8
= 31 +126= 157.\n\n**Final Answer**\n\boxed{
Rollout 0
rational number between 0 and 1 can be written as a fraction \(\frac{a}{b
Rollout 9
10 units. But in the square, that side EM is part of the square. Then the
Rollout 1
∈ S } are orthonormal. Hence, they can be considered as an orthonormal basis for
Rollout 1
non-empty.\n\nBut in infinite-dimensional spaces, you can have decreasing sequences of closed convex sets with empty intersection
Rollout 2
1000. When dealing with remainders, especially modulo 1000, it
Rollout 6
wait, the length log n + p(n) would vary for each n. For example, for different
Rollout 4
multiplied by 100%, I think. But first, I should probably find the relative errors in
Rollout 2
possible? The product is 9*99*999*... So thinking about modulus
Rollout 4
.8388, so total≈14,328 +477.8
Top 16 Negative Activations
Rollout 7
5)x\n\n-3P=(74 -A) x.\n\nWait, substitute P=2N
Rollout 7
/3) = (75*2)/3 *N=150/3 *N
Rollout 7
this math problem about participants in a test. Let me first try to understand what's being asked here.\n\n
Rollout 7
)x\n\n-3P=(74 -A) x.\n\nWait, substitute P=2N/
Rollout 1
2 - (2α - ||y||2) = 0 ⇒ (2c - d2
Rollout 9
: GE: y=(-1/5)(10)+2= -2 +2=0,
Rollout 1
)/2 - (2α - ||y||2) = 0 ⇒ (2c - d
Rollout 7
79P =79x - (A +5)x\n\n-3P=(74 -
Rollout 9
00 => y = 5.\n\nTherefore, the y-coordinate of point G must be 5,
Rollout 1
4 = d2 - (6d2/4) = d2 - 3)1.
Rollout 7
, the x terms:\n\n(75 -59)x=16x.\n\nTherefore, the equation
Rollout 7
+59*(N/3) -59x =71N\n\nCalculate the terms:\n\n
Rollout 8
5−r greens in order where all reds are adjacent and all greens adjacent. So the number of
Rollout 7
score becomes 66N + 5N = 71N. But also, the total
Rollout 3
=28,224.0\n\n168.5^2=168
Rollout 9
0 would be (10 *10)/2 =50, so overlap area is 5

Layer 50

GATE_PROJ

Top 16 Positive Activations
Rollout 4
. The equivalent capacitance for capacitors in series is given by:\n\n\[\nC_{\text{
Rollout 2
times (-1)^{997} \equiv 1 \times 3 \times (-1
Rollout 9
\(x = 0\) to \(x = 10\):\n \[\n \text
Rollout 8
400 + 1200 = 7200\).\n\nAdding both cases,
Rollout 5
} = \frac{2}{3}\n\]\nSolving this, we find:\n\[\n3
Rollout 8
= 7440\).\n\nThe probability is \(\frac{7440}{3
Rollout 5
midpoints of the legs, has a length equal to the average of the two bases, \( b +
Rollout 9
0}{h} = 80 \implies \frac{500}{h} =
Rollout 8
same color).\n - Total for Case 1: \(2 \times 5! = 2
Rollout 2
^1 - 1)(2^2 - 1) \times (-1)^{99
Rollout 7
. Total=186+486=672=56*12.
Rollout 9
500}{h}\n \]\n - Setting the area equal to 80:\n
Rollout 5
The areas of the regions divided by this midline are in the ratio 2:3. \n\nThe
Rollout 4
equivalent capacitance for capacitors in series is given by:\n\n\[\nC_{\text{eq}}
Rollout 8
Adding both cases, the total number of happy sequences is \(240 + 7200
Rollout 8
out 5 cards from 10, which is \(10 \times 9 \times
Top 16 Negative Activations
Rollout 3
161.4 m/s.\n\nAs an assistant, I need to perhaps suggest both possibilities, but
Rollout 3
m/s. But need to specify.\n\nAlternatively, here's another angle: gas thermometric speeds. At
Rollout 8
400 + 2400 + 1200 = (1200
Rollout 4
) method.\n\nAlternatively, check the question one more time. The question says: "What is the percentage
Rollout 1
imposed by each x in S are consistent.\n\nAnother thought: think of 2y as a functional.
Rollout 1
al through such a shift and scaling.\n\nSumming up, then despite the difficulty with coordinate system, using
Rollout 0
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number, write it as a
Rollout 1
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$
Rollout 3
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in the heavy
Rollout 4
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values $C
Rollout 6
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $L \
Rollout 7
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants is
Rollout 8
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red cards and
Rollout 9
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides of length
Rollout 5
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a trapezoid is
Rollout 4
2)) ] = (2000 * 15 )/(3000 *

UP_PROJ

Top 16 Positive Activations
Rollout 1
x1 - x2, 2y⟩ = ||x1||2 - ||x2||
Rollout 1
x and x',\n\n||x - x'||2 = d2,\n\nand ⟨x, x'⟩
Rollout 2
,000 -891=890,109. mod100
Rollout 2
9*9999≡109*999≡109* -1
Rollout 1
d, then:\n\n||x - x'||2 = d2,\n\nwhich is ⟨x - x',
Rollout 1
.\n\nBut using the inner product formula:\n\nd2 = ⟨x - x', x - x'⟩
Rollout 2
00 -891=890,109. mod1000,
Rollout 1
we can check:\n\n||x - x'||2 = ||(x - y) - (x'
Rollout 2
1). But (10^4 -1)=9999. 9999
Rollout 2
9\n\nk=3, so10^3 -1=999≡-1 mod
Rollout 9
ies \frac{500}{h} = 20 \implies h = 2
Rollout 2
9: 891*999=891*(1000 -1
Rollout 2
000 -109=108,891≡891 mod1
Rollout 1
S, then ⟨x - x', y⟩ = ⟨x, y⟩ - ⟨x', y
Rollout 1
⟨x - y, x' - y⟩ = ⟨x, x'⟩ - ⟨x,
Rollout 9
(0,50/h) and (0,10 -50/h). Since 5
Top 16 Negative Activations
Rollout 4
4)/2≈6 pF. Wait, but when we did the partial derivatives, we got ~
Rollout 0
8\), since each pair is counted twice. But actually, since we are considering ordered pairs where a
Rollout 3
people use 25°C as standard conditions. But in the absence of information, I think STP
Rollout 8
,5) if we have an unordered hand. However, since the order matters, the number of sequences
Rollout 7
suggests that perhaps the increase affects the promotion status? But the problem statement seems a bit ambiguous here. If
Rollout 4
is [0.003]^2. But wait:\n\nFirst term inside the formula is [ (
Rollout 9
and x=10 have different distance between. But in reality, these lines are not vertical but the
Rollout 9
the diagonal from E to M? Wait, no. Wait, E is bottom-right and M is top
Rollout 9
h is from 0 to 10? But EM is the vertical edge of the square. So
Rollout 3
a bit different from the root mean square speed. But maybe in this context, they just want the root
Rollout 4
0.003\n\nWait, no, actually. The term inside the sqrt for the first
Rollout 0
0! but 7^3 does not.\n\nBut exponent of 7 is 2, which is
Rollout 4
6, 2.6% total spread. But since normally uncertainty is defined as ± half the total
Rollout 9
=0 and x=10). Wait, but in trapezoid area formula:\n\nAverage of
Rollout 0
= 0), so that's exponent 2.\n\nWait, 7^2 divides into 2
Rollout 8
consider each color sequence as equally probable? Wait, but if the cards are distinguishable, then even if

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
for Rn, which is approx. 1/8 the speed (mass is ~8 times higher
Rollout 3
5.87 /337≈0.047.\n\nTherefore, sqrt≈16
Rollout 2
1mod125. 125*7=875, 891
Rollout 6
polynomial time given the input length (i.e., 1^n), then the advice isn't giving any
Rollout 3
5.87/337≈0.047. Hence, 168
Rollout 3
5.87)/337≈0.047. Thus, sqrt≈16
Rollout 2
1 mod125: 125*7=875. 891
Rollout 3
)=15.87/337≈0.047. Hence, 1
Rollout 9
For h=20, point G is at 10 -20 = -10,
Rollout 6
, but the problem doesn't say that k is given as part of the input. Wait, the input
Rollout 6
advice α_n. But encoding n can be done via binary representation. Then for each n, the number
Rollout 1
. However, Hilbert spaces are only locally compact in finite dimensions, thus only in finite dimensions would intersection
Rollout 6
s \in S$ and get the correct answer in return in constant time. [/*]\n \n
Rollout 3
(15.87)/337≈0.047. Thus, sqrt≈
Rollout 6
to one string of length m per n equal to p^{-1}(m). However, if p is
Rollout 6
, since A can compute α_n in poly-time given n, then in the machine M(x, α
Top 16 Negative Activations
Rollout 7
7)there is no solution. But the problem says "Find all possible values of N", so maybe
Rollout 3
temperature is required? But no, the problem says "compute the mean molecular speed," so they must want
Rollout 8
to count how many of these sequences have all red cards adjacent and all green cards adjacent.\n\nSo possible scenarios
Rollout 3
the temperature is required? But no, the problem says "compute the mean molecular speed," so they must
Rollout 7
But that's conflicting with the problem's request to "Find all possible values...", unless there are constraints I
Rollout 4
55%. So approximately 0.8776%.\n\nThus, the percentage error is approximately
Rollout 7
is no possible value of N.\n\nBut the problem states "Find all possible values ofN". If not
Rollout 7
the actual marks that count are the new ones. But the problem says passmark is fixed at65
Rollout 7
the pass mark changes. Wait, original problem states "the passmark is fixed at 65".
Rollout 8
R's, 5! card sequences.\n\nIf 1<=r<=4:\n\nThen, the color
Rollout 7
no possible value of N.\n\nBut the problem states "Find all possible values ofN". If not possible
Rollout 1
, x' in S?\n\nSuppose that all points in S lie on a sphere in H, then
Rollout 7
point. Let's read the original problem again:\n\n"due to an error in the wording of a question
Rollout 1
S with multiple points at distance d from 0 and from each other, such a y cannot exists.
Rollout 1
terms. Let me see.\n\nSuppose that all points in S lie in the set {x ∈ H
Rollout 6
the problem says "Given k ∈ N sparse sets S_1,...,S_k", so k is

Layer 51

GATE_PROJ

Top 16 Positive Activations
Rollout 5
+ x)) = 125H\n\nLet me factor H out:\n\n(x + 17
Rollout 5
175)/(x + 75)\n\nLet me write the entire equation:\n\n(x + 1
Rollout 5
75) = 125\n\nLet me let y = x + 75.
Rollout 5
x)) = 125H\n\nLet me factor H out:\n\n(x + 175
Rollout 9
, maybe visualizing coordinates would help. Let's place EM as the vertical segment from (10,
Rollout 4
15 )^2\n\nHmm, let's compute these step by step.\n\nFirst, compute the
Rollout 9
,10). So these are straight lines. Let me find equations for GE and GM.\n\nEquation
Rollout 5
75)/(x + 75)\n\nLet me write the entire equation:\n\n(x + 17
Rollout 5
b = 175, height H, area S = (75 + 175
Rollout 9
0). So these are straight lines. Let me find equations for GE and GM.\n\nEquation for GE
Rollout 5
75) = 125\n\nLet me let y = x + 75. Then
Rollout 8
40/30240= Let's check by dividing numerator and denominator by 24
Rollout 4
2*(5.02)^2. Let's calculate.\n\nFirst, (5.02)^
Rollout 8
40 divided by 240. Let's see: 240×30 =
Rollout 4
1) is 0.01. Let's approximate:\n\n0.000076
Rollout 4
0 / 5025 �� let's divide 6,060,15
Top 16 Negative Activations
Rollout 4
energy stored in this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to
Rollout 2
is \(\boxed{109}\).<|im_end|>
Rollout 9
\) is \(\boxed{25}\).<|im_end|>
Rollout 0
!$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out
Rollout 0
is \(\boxed{128}\).<|im_end|>
Rollout 3
(Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the
Rollout 9
in $\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this
Rollout 3
on (Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate
Rollout 9
EM$ in $\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to
Rollout 8
Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need
Rollout 5
2/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need
Rollout 3
\, \text{m/s}\n\]<|im_end|>
Rollout 9
Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(AIME\) has sides of length
Rollout 1
y \text{ exists as described}}\n\]<|im_end|>
Rollout 5
x^2/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see.
Rollout 1
$ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show

UP_PROJ

Top 16 Positive Activations
Rollout 5
exceeding x2/100. \n\nFirst, let me recall some trapezoid properties.
Rollout 3
molecular speed of radon gas. Hmm, first, I remember there's a formula for the root mean
Rollout 1
cal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there exists
Rollout 1
/ sqrt{1 - a^, but this is stretching.\n\nBut since in each pair x, x
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this math
Rollout 5
100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve
Rollout 8
m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to find
Rollout 1
radius d / sqrt(2) must intersect.\n\nBut even in three dimensions, the intersection of three such
Rollout 9
triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry problem
Rollout 4
:\n\n2010 * 3015 = Compute 2000*30
Rollout 1
/ sqrt(2) centered at each point.\n\nBut in R^3, let's compute. Let
Rollout 9
requires calculating the region inside both.\n\nHmm. Let's visualize that. The triangle has a vertex outside the
Rollout 9
the square requires calculating the region inside both.\n\nHmm. Let's visualize that. The triangle has a vertex
Rollout 4
0.02 V. Hmm, alright, let's start by recalling how to calculate the equivalent
Rollout 2
000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here: I
Rollout 4
2 V. Hmm, alright, let's start by recalling how to calculate the equivalent capacitance of capac
Top 16 Negative Activations
Rollout 3
s2·mol). Then, dividing by M in kg/mol, the mol unit cancels, then
Rollout 3
*R*T)/(π*M)), R is in J/(mol·K), which is same as (kg
Rollout 3
m2)/(s2·mol) divided by (kg/mol) is (m2/s2).
Rollout 3
J/(mol·K), which is same as (kg·m2)/(s2·mol·
Rollout 3
·m2)/(s2·mol) divided by (kg/mol) is (m2/s2
Rollout 8
the probability is 31/126, so m=31, n=12
Rollout 8
). But the way the problem is phrased: "all the red cards laid out are adjacent and
Rollout 8
m=31, n=126.\n\nWait, but let's confirm. Wait, when
Rollout 7
12:\n\nA -42=56\n\nThus,A=98\n\nWait, but A
Rollout 7
b) impossible, no solutions. But language problems say "Find all possible values ofN". If "
Rollout 7
24,36. However, the problem states N<40. 36 is less
Rollout 3
.\n\nUse R=8.314 J/(mol·K). Correct.\n\nSpeed in m/s
Rollout 3
cancels, then kg in denominator; thus overall (kg·m2)/(s2·mol)
Rollout 4
statistical error propagation. The question is, when they say percentage error in the calculation, is it the maximum
Rollout 8
Wait, wait, no. Wait, the problem says "shuffles the 10 cards and lays
Rollout 3
))\n\nR = 8.314 J/(mol·K)\n\nAssume T = 2

DOWN_PROJ

Top 16 Positive Activations
Rollout 0
), it depends. As with N! for N>1.\n\nWait, for N >=2, N
Rollout 5
. Probably, I made an algebraic mistake.\n\nStarting from the two equations:\n\n1. (75
Rollout 0
0 and 1 with product a*b=720, gcd(a,b)=1, a <
Rollout 1
, since each sphere is closed and convex, and in a Hilbert space, the intersection of any number
Rollout 0
, each prime factor in 20! must go entirely to \(a\) or entirely to \(b
Rollout 8
31/126≈0.246.\n\nTo verify if that's correct.\n\n
Rollout 6
P/poly is equivalent to languages decidable by a polynomial-time Turing machine with a polynomial-length advice string
Rollout 1
use the following theorem: Any set of vectors in a Hilbert space can be isometrically embedded into
Rollout 1
spaces, you can have decreasing sequences of closed convex sets with empty intersection even if finite intersections are non-empty
Rollout 1
states that if we have a family of closed convex sets where every finite intersection is non-empty, then the
Rollout 1
. But in general, this requires the family to have the finite intersection property. So if every finite sub
Rollout 5
. Therefore, final answer is 181. \n\nI think that's correct. Let me ensure
Rollout 0
each prime factor in 20! must go entirely to \(a\) or entirely to \(b\
Rollout 1
, you can have decreasing sequences of closed convex sets with empty intersection even if finite intersections are non-empty.
Rollout 1
centered" (has the finite intersection property) and maybe in a weakly compact set. But in
Rollout 1
since each sphere is closed and convex, and in a Hilbert space, the intersection of any number of
Top 16 Negative Activations
Rollout 2
impossible because the numbers are huge. So, perhaps there's a pattern or periodic behavior we can exploit when
Rollout 5
/100. \n\nFirst, let me recall some trapezoid properties. In a tr
Rollout 1
radius d / sqrt(2} approx 0.707d. So sum of radii
Rollout 6
n + c, we can invert to n �� 2^{m - c}. The number of
Rollout 1
we need to find t and C such that all points in S satisfy that equation.\n\nAlternatively, writing this
Rollout 8
, GG RRR\n\nk = 4: RRRR G, G RRRR\n\nSo
Rollout 1
two distinct vectors are orthogonal.\n\nFirst, let me recall that an orthonormal system requires two things:
Rollout 4
to C_eq and V squared. Therefore, exponents are 1 for C_eq and 2 for
Rollout 2
1000. Hmm, okay, let's unpack this step by step.\n\nFirst, the problem
Rollout 7
12 +54N/4.\n\nConvert to same denominator:\n\n62N/12 +
Rollout 1
Alternatively, using the hypothesis that the mutual distances between points are all d, this might impose that the spheres
Rollout 4
.7749e-3*sqrt(10). Because sqrt(10)=3
Rollout 4
.02 V. Hmm, alright, let's start by recalling how to calculate the equivalent capacitance
Rollout 3
check online, but since I can't do that, I need to recall. In some problems, if
Rollout 1
y is the center of a sphere where all the points x in S lie on the sphere's surface,
Rollout 1
by a point \( y \) such that all points lie on a sphere centered at \( y \)

Layer 52

GATE_PROJ

Top 16 Positive Activations
Rollout 2
999 \times \cdots \times \underbrace{99\cdots9
Rollout 2
times \cdots \times \underbrace{99\cdots9}_{\text{9
Rollout 1
x, x' \in S \), the vectors \( \frac{\sqrt{2}}{d
Rollout 8
\(k = 2\): \(2 \times (5 \times 4 \times 5
Rollout 2
9 \times \cdots \times \underbrace{99\cdots9}_{\text
Rollout 8
k = 2\): \(2 \times (5 \times 4 \times 5 \
Rollout 0
case \(a = b\) (which is impossible since 20! is not a square).\n\n4
Rollout 2
999 \times \cdots \times \underbrace{99\cdots9
Rollout 1
\in S \), the vectors \( \frac{\sqrt{2}}{d}(x - y
Rollout 0
airs \(a < b\)**:\n - Since \(a\) and \(b\) are copr
Rollout 0
= b\) (which is impossible since 20! is not a square).\n\n4. **Result
Rollout 2
\times \cdots \times \underbrace{99\cdots9}_{\text{
Rollout 4
Delta C_{\text{eq}} = \sqrt{(0.36 \cdot 10
Rollout 2
\times \underbrace{99\cdots9}_{\text{999 9
Rollout 3
at standard temperature and pressure, with boiling point at -61.7 °C, so at ST
Rollout 8
- Calculations:\n - \(k = 1\): \(2 \times (5 \
Top 16 Negative Activations
Rollout 9
I'm confused. Wait squares have four sides. Let me actually assign coordinates to the square to make this
Rollout 9
. \n\nWait, in the square AIME, let me confirm the positions. If the square is labeled
Rollout 3
the formula applies. So, with that said, let me proceed. Let me check again.\n\nIf R
Rollout 3
around 170 m/s. Alternatively, let's check some resource. Wait, can't access
Rollout 2
≥3.\n\nSo similar to modulo8 analysis? Let's be precise.\n\nbreak down the terms in the
Rollout 3
is a heavy gas, so maybe at room temperature?\n\nBut surely, since no temperature is given, we
Rollout 3
So, with that said, let me proceed. Let me check again.\n\nIf Rn-22
Rollout 3
, perhaps there's a standard temperature to assume. Let me check online, but since I can't do
Rollout 5
, 181 is the correct answer. \n\n**Final Answer**\n\boxed{181
Rollout 8
we can calculate the number of such happy sequences. Let's consider two cases:\n\nCase 1: All
Rollout 3
compute it, maybe we need to state standard assumptions.\n\nAlternatively, maybe the term "mean molecular speed"
Rollout 3
is taken as 300 K for approximation.\n\nWait, given that no temperature is given, perhaps
Rollout 3
temperature, which is around 298 K. Maybe. Alternatively, as room temperature sometimes is taken
Rollout 3
.\n\nBut then again, maybe even room temperature. Let me see. Let's do a quick estimation.
Rollout 3
I think STP is 0°C. But let's see if it matters—wait, if the
Rollout 3
maybe in the original context, the temperature is standard. Maybe just proceed with 298 K and

UP_PROJ

Top 16 Positive Activations
Rollout 0
2,3,5,7 are even?\n\nWait floor operations:\n\nFor prime 2:
Rollout 9
5)/(10 - (10 - h)) = 5/h.\n\nTherefore, equation is y
Rollout 0
(even), 2 (7) (even?), no 7 has exponent 2? Wait for
Rollout 9
)/(10 - (10 - h)) = 5/h.\n\nTherefore, equation is y -
Rollout 1
2 - 2c + d2)/2 = 0 ⇒ 0 = 0.\n\nThus
Rollout 4
14.8058)/15 = 0.39/15 ��
Rollout 9
)/(10 - (10 - h)) = (-5)/h.\n\nThe equation is y -
Rollout 0
(even?), no 7 has exponent 2? Wait for 7: floor(20/
Rollout 0
4,7:2, which are all even? Wait, 18,8,4,
Rollout 5
30625)/2] = sqrt(36250/2) =
Rollout 5
2500 = 125y\n\nSimplify:\n\ny^2 - 25
Rollout 5
125x + 9375\n\nSimplify:\n\nx^2 + 12
Rollout 9
5)/(10 - (10 - h)) = (-5)/h.\n\nThe equation is y
Rollout 7
213N -209N)/3=4N/3.\n\nTherefore:\n\n16
Rollout 5
25 + 30625)/2] = sqrt(36250/
Rollout 7
213N -205N)/3=8N/3.\n\nHence:\n\n3
Top 16 Negative Activations
Rollout 2
seems impossible because the numbers are huge. So, perhaps there's a pattern or periodic behavior we can exploit
Rollout 1
x'||2 - d2)/2.\n\nBut perhaps another way is expressing this by defining various terms.
Rollout 1
infinite set where each pair is separated by d, perhaps S must lie on such a quadratic manifold. But
Rollout 1
that this holds for all x in S.\n\nBut maybe we can see this as a functional equation. Let
Rollout 1
't necessarily contain a linear subspace.\n\nAlternatively, perhaps we can impose some structure. Let me fix k
Rollout 1
how can we construct such a y?\n\nAlternatively, perhaps by considering properties of such sets S. If S
Rollout 1
able, so traditional averaging may not work. But perhaps we can use completeness of the Hilbert space.\n\n
Rollout 2
product. But before multiplying everything out, maybe see if we can find when (2^k -1
Rollout 1
must simultaneously satisfy for each xi in S.\n\nBut perhaps all these equations can be combined into one, with
Rollout 1
, but we already new that.\n\nGiven that, perhaps the only way to proceed is to conclude that such
Rollout 1
appropriately could make them orthonormal.\n\nAlternatively, maybe in our case, if we shift the origin to
Rollout 1
d2/2)} / 2.\n\nBut perhaps it's hard.\n\nWait, back to the equation
Rollout 2
the Chinese Remainder Theorem. Let me check if that's feasible.\n\nStarting with modulo 8:
Rollout 2
this directly seems impossible because the numbers are huge. So, perhaps there's a pattern or periodic behavior we
Rollout 1
of modified hyperplane equation with a quadratic term.\n\nBut given that S is an infinite set where each pair
Rollout 6
blow up the length too much. For example, maybe for each x in S_i, encode it in

DOWN_PROJ

Top 16 Positive Activations
Rollout 2
91 mod125: 125*7=875. 89
Rollout 2
91mod125. 125*7=875, 89
Rollout 9
Wait, perhaps my assumption is incorrect. Let's recast this. I need to check whether when h
Rollout 3
with precise calculation using more decimal places.\n\nLet me recalculate the exact value.\n\n8 * 8.
Rollout 4
025:\n\nFirst, 5025 * 1200 = 6,
Rollout 0
3, 17, 19 each have exponent 1 in 20!.\n\nSo ex
Rollout 0
product a*b=20! means that each prime's exponent in 20! is divided between
Rollout 4
02 *10, etc. Let's recalculate the derivatives values more precisely:\n\nStarting with Δ
Rollout 0
which it is not (since exponents in its prime factorization are all even? Wait, 2
Rollout 4
5025 * 1200 = 6,030,000
Rollout 0
n! is not a square (unless for example primes with even exponents), it depends. As with
Rollout 0
in a product a=144, b=5. To reach a < b, the valid
Rollout 3
with ~168 m/s. So this checks out.\n\nSo all steps are correct.\n\nThus,
Rollout 5
x squared is 25*725 = 18125. Then x2
Rollout 0
squareful; but even so, does assigning a prime to a versus b affect the magnitude of a compared
Rollout 8
40. Let's see: 240×30 =7200,
Top 16 Negative Activations
Rollout 3
(225,298)= approx 474.6 m/s. So about
Rollout 3
Then sqrt(225,298)= approx 474.6 m/s.
Rollout 3
28,408.12)= 168.5\n\nThus, answer is
Rollout 3
(28408.12)= 168.546... Hence,
Rollout 3
Let's see.\n\n168^2 = (170 - 2)^2 =
Rollout 3
225,753 ) �� 475 m/s. Which is correct.\n\n
Rollout 3
0\n\nsqrt(27,950)= approximately 167.2 m/s.
Rollout 3
,408.12) �� 168.5 m/s\n\nSo approximately
Rollout 3
0; sqrt(27,950)=167.2 m/s. So around
Rollout 5
equation 1: h = (125H)/(75 + x). Plug into equation
Rollout 4
.5%), and 0.02V (0.04%), so maybe a couple of
Rollout 3
sqrt(27,950)= approximately 167.2 m/s. So ~
Rollout 3
(28,433) �� 168.6 m/s.\n\nSo my
Rollout 5
the first equation: h = (125H)/(75 + x)\nFrom the second equation
Rollout 3
68.5)=15.87/337≈0.047.
Rollout 3
8,436). Let's see, 168^2 = 28,

Layer 53

GATE_PROJ

Top 16 Positive Activations
Rollout 8
*9. 31 and 126 are coprime, so yeah, simplified.\n\n
Rollout 9
\n\nSo 20y = 100 => y = 5.\n\nTherefore, the y
Rollout 8
30240.\n\nSimplify this fraction.\n\nFirst, divide numerator and denominator by 24
Rollout 9
But the problem says this area is 80. Therefore:\n\n100 -500/h
Rollout 8
30240. But let me verify that. Yes, 10 choices for first
Rollout 8
.\n\nTherefore, 31/126. Then check if they can be simplified. Since
Rollout 7
2N)/3, R = N/3.\n\nSubstitute P and R:\n\n75*( (
Rollout 8
26. Then check if they can be simplified. Since 31 is prime. 12
Rollout 8
10P5=30240. Then 7440/302
Rollout 5
\n\nTherefore, 181 is the correct answer. \n\n**Final Answer**\n\boxed{1
Rollout 8
440 / 30240.\n\nSimplify this fraction.\n\nFirst, divide numerator and
Rollout 9
square, so the triangle is completely within the square. Wait, because the square spans from x=0
Rollout 2
ime, m must be ��0 mod8. Therefore, m=8n for some integer n
Rollout 9
/h)x + (50/h) +0? Wait, let's verify:\n\nFrom G(1
Rollout 4
2 * 0.004)^2 )\n\nCalculating 2*0.004
Rollout 8
10P5=30240. 7440/3024
Top 16 Negative Activations
Rollout 5
of the legs divides the trapezoid into two regions whose areas are in the ratio $2:
Rollout 8
and $5$ green cards. She shuffles the $10$ cards and lays out $5
Rollout 8
cards and $5$ green cards. She shuffles the $10$ cards and lays out $
Rollout 3
921;162^2=26,244. So it's between
Rollout 6
bb{N} \mapsto \mathbb{N}$ such that for each $n
Rollout 3
12 * 270 = 17,958.24, plus
Rollout 3
4. 169^2 = 28,561. So sqrt(2
Rollout 8
but RRRGR will not. The probability that Kathy will be happy is $\frac{m}{n
Rollout 8
shuffles the $10$ cards and lays out $5$ of the cards in a row in
Rollout 3
2 * 270 = 17,958.24, plus 6
Rollout 3
0.222 \approx 0.697433\n\]\n\
Rollout 3
.222 �� 0.697433\n\nThus, the division:
Rollout 6
{bad-angel}} \) by setting \( A(n) \) to output a null advice string
Rollout 8
. She will be happy if and only if all the red cards laid out are adjacent and all the green
Rollout 3
.942*298 ��24.942*300=
Rollout 6
bad-angel}} \) does not provide the nondeterminism inherent in \( \textbf{NP

UP_PROJ

Top 16 Positive Activations
Rollout 4
0.003605^2 = approx 0.000013
Rollout 4
6055555)^2= 0.00001300
Rollout 3
,820.576 �� 0.697433 ��
Rollout 5
2:\n\n(x + 175)(H - (125H)/(75 + x
Rollout 6
one and keep the combined set sparse? Let's think. Suppose each S_i has at most p(n
Rollout 4
_eq is ΔC_eq / C_eq = sqrt( [ (C2 / (C1 (C
Rollout 5
2:\n\n(x + 175)(H - (125H)/(75 +
Rollout 5
So:\n\n(x + 175)(H - (125H)/(75 + x
Rollout 3
kg/mol.\n\nNow, the average speed is sqrt(8RT/(πM)).\n\nBut if the
Rollout 3
.512\n\nThen, 66.512 * 298:\n\nLet
Rollout 4
36055555)^2= 0.0000130
Rollout 3
28 g/mol, the average speed is sqrt(8*8.314*29
Rollout 3
.0\n\n168.5^2=168^2 + 2*1
Rollout 4
.008)^2 ) �� sqrt(0.000013 +
Rollout 3
�� 3.1416 * 0.222 �� 0.
Rollout 4
formula for the uncertainty ΔC_eq would be sqrt( (��C_eq/∂C1 * Δ
Top 16 Negative Activations
Rollout 2
**Final Answer**\n\boxed{109}\n\n<|im_start|>answer\nTo find the remainder when \(9
Rollout 9
.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(AIME\) has sides
Rollout 5
**Final Answer**\n\boxed{181}\n\n<|im_start|>answer\nLet the shorter base be \( b
Rollout 7
boxed{24}, \boxed{36}\n\n(b) No solution exists.\n\n<|im_start|>answer\nGiven
Rollout 8
**Final Answer**\n\boxed{157}\n\n<|im_start|>answer\nKathy has 5 red cards
Rollout 0
**Final Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven a rational number between 0
Rollout 8
2: RRGGG, GGGGRR\n\nWait, wait, maybe I need to reorgan
Rollout 8
= 157.\n\n**Final Answer**\n\boxed{157}\n\n<|im_start|>answer\nK
Rollout 1
a point } y \text{ exists as described}}\n\n<|im_start|>answer\nGiven an infinite-dimensional Hilbert space
Rollout 0
1}=128.\n\n**Final Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven
Rollout 0
2^{8-1}=128.\n\n**Final Answer**\n\boxed{128}\n\n
Rollout 5
181 is the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n
Rollout 5
is 181. Therefore, the answer is 181. \n\nWait, but let
Rollout 2
boxed answer is \boxed{109}\n\n**Final Answer**\n\boxed{109}\n\n
Rollout 9
the altitude being h=25.\n\nPerfect.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>
Rollout 7
36}\).\n\n(b) No solution exists.<|im_end|>

DOWN_PROJ

Top 16 Positive Activations
Rollout 2
109 mod125.\n\nSo summary so far:\n\nP≡5 mod8\n\nand P
Rollout 2
=3 onward are -1 mod8. So starting from k=3 up to k=99
Rollout 7
-54) = 2/3\n\nMultiply both sides by 12:\n\nA -5
Rollout 4
{0.88\%}.\n\n<|im_start|>answer\nTo find the percentage error in the calculation of
Rollout 3
.028) ) �� sqrt( (19872.7 ) / (
Rollout 1
⟩ = k for all x ∈ S.\n\nThis is a system of equations. For all x in S
Rollout 7
74 -A)*(N/12)\n\nMultiply both sides by12:\n\n-24N
Rollout 5
+ x)) = 125H\n\nMultiply out:\n\n(x + 175)H
Rollout 2
891*(1000 -1)=891*1000 -8
Rollout 2
891*(1000 -1)=891,000 -89
Rollout 3
0.222)) �� sqrt( (8*8.314*29
Rollout 5
100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve
Rollout 2
not coprime. So 10 is not invertible mod125. However, perhaps
Rollout 1
x, x⟩ + ⟨x', x'⟩ - d2)/2.\n\nTherefore, substituting
Rollout 8
also not work. So we need to calculate the number of such "happy" sequences divided by the total
Rollout 5
)]/y) = 125\n\nMultiply through by y to eliminate the denominator:\n\ny(y
Top 16 Negative Activations
Rollout 3
. At STP, the molar volume is 22.4 liters, which is for ideal
Rollout 3
8,436). Let's see, 168^2 = 28,
Rollout 3
applies. So, with that said, let me proceed. Let me check again.\n\nIf Rn-
Rollout 2
is the case?\n\nWait, but this conflicts with our result for modulo8. If P≡5 mod
Rollout 2
0^{k-3}≡0*10^{k-3}=0 mod100
Rollout 0
,3,5,7 are even?\n\nWait floor operations:\n\nFor prime 2: 20
Rollout 0
9 have exponent 1.\n\nSo primes 2: 18, 3: 8,
Rollout 3
,408.12} \approx 168.5 \, \text{
Rollout 7
the original scores. But this is ambiguous.\n\nAn alternative interpretation is that when they increased all scores by5
Rollout 6
Because in the problem statement, given S_1,...,S_k (finite k), we need to
Rollout 3
original context, the temperature is standard. Maybe just proceed with 298 K and note the assumption
Rollout 3
222 kg/mol. Correct.\n\nUse R=8.314 J/(mol·K
Rollout 3
compute speeds. So, without further information, to proceed with room temperature 25°C (29
Rollout 0
2 to a, then a would have 2 to the 18th power, whereas
Rollout 9
��[0 to10] [ (2/5 x +6 ) ] dx\n\n=
Rollout 4
when we did the partial derivatives, we got ~4.327 pF. But here,

Layer 54

GATE_PROJ

Top 16 Positive Activations
Rollout 4
0.02 / 5.00 = 0.004.\n\nNow, since
Rollout 4
327 / 1200 �� 0.003606.
Rollout 8
31 +126=157.\n\nBut I want to cross-verify.\n\nWait
Rollout 4
the percentage error is approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation
Rollout 4
.327 / 1200 = let's compute that more accurately.\n\n4.3
Rollout 4
.003606. So, approximately 0.36%.\n\nNow, moving on
Rollout 4
.327 / 1200 �� 0.003606
Rollout 4
= 10 / 2000 = 0.005. Similarly, for
Rollout 5
5 is 181. Therefore, final answer is 181. \n\nI think that
Rollout 4
be 0.195 /15 =0.013, 1.3
Rollout 8
the probability is 31/126, so m=31, n=12
Rollout 5
, 181 is the correct answer. \n\n**Final Answer**\n\boxed{181
Rollout 4
0.008774, or 0.8774%\n\nThus,
Rollout 9
is \boxed{25} . Seems done.\n\nThat's logical. Another way:\n\nFrom h=
Rollout 7
=49/6≈8.1667\n\nThus:\n\nA=12*(4
Rollout 8
31 +126= 157.\n\n**Final Answer**\n\boxed{15
Top 16 Negative Activations
Rollout 6
N}}$ with $\alpha_n \in \{0,1\}^{p(n
Rollout 3
and pressure, with boiling point at -61.7 °C, so at STP (0
Rollout 3
pi*0.028= ~0.08796\n\n19,8
Rollout 2
Find the remainder when $9 \times 99 \times 999 \times \cd
Rollout 5
two regions whose areas are in the ratio $2: 3$ . Let $x$ be the
Rollout 8
<|im_start|>user\nKathy has $5$ red cards and $5$ green cards. She shuffles
Rollout 0
to be coprime. Assign each prime of 20! to either a or b. So
Rollout 4
5000^2} = 0.16\n\]\n\nThe uncertainties are:\n\n\
Rollout 1
Space Properties**:\n - In an infinite-dimensional Hilbert space, one can have uncountable or
Rollout 6
}}$ with $\alpha_n \in \{0,1\}^{p(n)}
Rollout 3
: Calculate the average speed of N2 molecules at 25°C. And they do so using sqrt
Rollout 6
$ such that for every $x \in \{0,1\}^n$
Rollout 3
222 g/mol, which is 0.222 kg/mol. Let's check once
Rollout 1
subset of \( S \), the intersection of the corresponding spheres is non-empty because the mutual distances and radi
Rollout 4
(4.98)^2=24.8004\n\nE_min=0.
Rollout 3
/(πM)). Similarly, it shows that at 25°C (298 K), average

UP_PROJ

Top 16 Positive Activations
Rollout 5
)) = 125H\n\nLet me factor H out:\n\n(x + 175)
Rollout 7
18N =56N/3\n\nFactor x:\n\nx(A -54) +1
Rollout 5
5 + x) = 125\n\nFactor the second term:\n\nLeft side = (x +
Rollout 5
75 + x)\n\nExpand the left-hand side:\n\nx^2 - 50x +
Rollout 7
4) = 2N/3\n\nDivide both sides by N (assuming N ≠0,
Rollout 9
^2 + (10 - y)^2]\n\nSquaring both sides:\n\nh^2 +
Rollout 5
(75 + x)\n\nExpand the left-hand side:\n\nx^2 - 50x +
Rollout 7
66P + 66R\n\nSubtracting 66P and 56R
Rollout 7
6(P + R)\n\nLet me simplify this equation:\n\n71P + 56R =
Rollout 5
0 = 125y\n\nBring all terms to left side:\n\ny^2 - 2
Rollout 7
= 66(P + R)\n\nLet me simplify this equation:\n\n71P + 56
Rollout 5
00) = 125y\n\nFactor left-hand side:\n\n(y + 100
Rollout 7
R = 66P + 66R\n\nSubtracting 66P and
Rollout 7
x)=56*(N/3)\n\nSimplify:\n\nA*x +54*N/3 -
Rollout 5
= 125y\n\nBring all terms to left side:\n\ny^2 - 25
Rollout 5
x)] = 125H\n\nDivide both sides by H (assuming H ≠ 0
Top 16 Negative Activations
Rollout 8
m = 31\) and \(n = 126\). Thus, \(m +
Rollout 0
6}{2} = 128\).\n\nThus, the number of rational numbers between 0
Rollout 9
h}x + 10 - \frac{50}{h}\).\n\n5. **Intersection
Rollout 1
formed by translating and scaling the set \( S \).\n\n\[\n\boxed{\text{Such a point
Rollout 9
EM\) in \(\triangle GEM\).\n\n1. **Vertices of the Square**:\n -
Rollout 9
implies h = 25\n \]\n\nThus, the length of the altitude to \(EM
Rollout 8
, \(m + n = 31 + 126 = 157\).\n\n
Rollout 9
\frac{500}{h} = 80 \implies \frac{50
Rollout 0
, the number of rational numbers between 0 and 1 for which the product of the numerator and denominator
Rollout 2
\mod 1000\n \]\n\nThus, the remainder when the product is divided by
Rollout 8
126 = 157\).\n\n\[\n\boxed{157}\n\
Rollout 0
a \times b = 20!\).\n\n1. **Prime Factorization of 20!
Rollout 7
possible values of \( N \) satisfy the conditions.\n\n### Final Answers\n\n(a) The possible values of
Rollout 3
text{Denominator: } \pi \times 0.222 \approx 0.
Rollout 9
\(EM\) in \(\triangle GEM\) is \(\boxed{25}\).<|im_end|>
Rollout 2
the product is divided by 1000 is \(\boxed{109}\).<|im_end|>

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
� 28,436. Then take square root.\n\nsqrt(28,43
Rollout 4
.00007696\n\nTake the square root: sqrt(0.00
Rollout 3
,408.12\n\nNow, take the square root of 28,40
Rollout 3
4, as 161^2=25,921;162^
Rollout 3
921;162^2=26,244. So it's between
Rollout 4
percentage error is approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of
Rollout 4
2 = 0.00001296\n\nAdding those together: 0.
Rollout 4
4)^2 )=sqrt(0.1296 +0.64)=sqrt(
Rollout 4
max-min approach assumes that all errors contribute in the same direction, which is more of a worst-case scenario
Rollout 0
a^2 < 20! or a < sqrt(20!). Therefore, the number of
Rollout 0
2^{8-1}=128.\n\n**Final Answer**\n\boxed{128}\n\n
Rollout 4
is approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of the energy
Rollout 8
+126= 157.\n\n**Final Answer**\n\boxed{157}\n\n
Rollout 5
181 is the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n
Rollout 9
. If moving around square AIME, probably vertices in order A-I-M-E to make AIME.
Rollout 1
y||2 - d2/2, we can express 2α = c + ||y||
Top 16 Negative Activations
Rollout 5
base. The segment that joins the midpoints of the legs divides the trapezoid into two regions
Rollout 1
we have a set of vectors with equal norms and their differences have equal norms, this often relates to them
Rollout 0
the problem reduces to finding the number of coprime pairs \((a, b)\) where \(
Rollout 1
use that in H, you can have orthonormal systems of arbitrary cardinality.\n\nSo, using Z
Rollout 1
Hilbert space, one can have an orthonormal basis of arbitrary cardinality or an appropriate system,
Rollout 1
in non-separable Hilbert spaces, you have uncountable orthonormal bases, while in separ
Rollout 1
need the family to be "centered" (has the finite intersection property) and maybe in a weak
Rollout 0
both.\n\nTherefore, the number of such coprime pairs is \(2^k\), where \(
Rollout 0
20!. So they form a coprime pair whose product is 20!. Such pairs
Rollout 1
caled. Thus, this becomes, an orthonormal family. So in infinite dimensions, such a family
Rollout 1
certain conditions hold, such as the family of spheres has the finite intersection property.\n\nBut the problem says H
Rollout 1
but in a Hilbert space, any orthonormal system must be countable because the space is separ
Rollout 4
62 +0.0082)=sqrt(0.0000129
Rollout 1
" or "orthonormal basis" even if uncountable. However, typically in non-separable
Rollout 1
.\n\nAlternatively, since you can extend any orthonormal system to a larger one, again the infinite dimension
Rollout 0
8 primes. Therefore, the number of coprime pairs would be \(2^8 = 2

Layer 55

GATE_PROJ

Top 16 Positive Activations
Rollout 4
is approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of the energy
Rollout 4
[\n\boxed{0.88\%}\n\]<|im_end|>
Rollout 3
69} \, \text{m/s}\n\]<|im_end|>
Rollout 3
, \text{m/s}\).\n\n**Final Answer:**\n\[\n\boxed{169
Rollout 8
m = 31\) and \(n = 126\). Thus, \(m +
Rollout 4
) using error propagation. The partial derivatives are:\n\n\[\n\frac{\partial C_{\text{
Rollout 9
h}x + 10 - \frac{50}{h}\).\n\n5. **Intersection
Rollout 1
such spheres is non-empty.\n\n5. **Conclusion**:\n - The existence of such a point \(
Rollout 4
_{\text{eq}} \) is:\n\n\[\n\frac{\Delta C_{\text{
Rollout 3
\text{m/s}\).\n\n**Final Answer:**\n\[\n\boxed{169}
Rollout 8
of the same color).\n - Total for Case 1: \(2 \times 5! =
Rollout 1
a point } y \text{ exists as described}}\n\]<|im_end|>
Rollout 3
Answer**\nThe mean molecular speed of radon gas is \boxed{169} m/s.\n\n
Rollout 3
answer 169 m/s.\n\n**Final Answer**\n\boxed{170} m/s
Rollout 4
0^2} = 0.16\n\]\n\nThe uncertainties are:\n\n\[\n\Delta
Rollout 0
, 13, 17, and 19. These are 8 distinct primes.\n\n
Top 16 Negative Activations
Rollout 9
ating GE distance:\n\nGE is from G (10 - h, y) to E (10
Rollout 9
h, and the location of G is (10 - h, y). But the triangle being is
Rollout 7
(but <65) and (N/3 -x) students with original scores <60
Rollout 8
sequences is 2 * (P(5,2)*P(5,3))=2*
Rollout 9
10,10)\nG is (10 - h, y), but we need to determine
Rollout 9
the square (to the left), at (10 - h,5) where h >10
Rollout 9
base EM. Therefore, G is at (10 - h, 5). The altitude is h
Rollout 7
N/3 -x) = (N/3 -N/12)= (4N -
Rollout 9
Distance GE = sqrt[(10 - (10 - h))^2 + (0 - y)^
Rollout 9
0, then the point G is at (10 - h,5), which would be inside the
Rollout 9
Wait, let's verify:\n\nFrom G(10 - h,5) to E(10
Rollout 9
.\n\nEquation for GE (from G(10 - h,5) to E(10
Rollout 9
2]\n\nSimilarly, GM is from G (10 - h, y) to M (10
Rollout 9
=(10,10), G=(10 - h,5). Because we found earlier that
Rollout 7
.\n\nThen, T2 = B*(N/3 -x).\n\nSo,\n\nA*x + B*(
Rollout 9
x=0, y = (-5/h)(0) + (50/h) = 5

UP_PROJ

Top 16 Positive Activations
Rollout 2
Modulo 125 Calculation:**\n - Each term \(10^k - 1
Rollout 2
is \(2^k - 1\).\n - For \(k \geq 3\
Rollout 9
G\) to \(EM\) be \(h\). The coordinates of \(G\) are \((1
Rollout 4
.16\n\]\n\nThe uncertainties are:\n\n\[\n\Delta C_{\text{eq}}
Rollout 1
. **Orthonormal Conditions**:\n - For each \( x \in S \), the vector
Rollout 7
a)\n\n1. **Original equations:**\n - Total score: \( 66N =
Rollout 6
\) with the binary representation of \( i \). Since each \( S_i \) is sparse,
Rollout 5
\implies b = 75\n\]\nThus, the bases are 75 and
Rollout 6
). Since each \( S_i \) is sparse, for each length \( n \), the number of
Rollout 1
1. **Orthonormal Conditions**:\n - For each \( x \in S \), the
Rollout 2
results using the Chinese Remainder Theorem.\n\n1. **Modulo 8 Calculation:**\n -
Rollout 1
d}{\sqrt{2}} \).\n - For any two distinct \( x, x' \
Rollout 2
\(2^k - 1\).\n - For \(k \geq 3\),
Rollout 6
of \( i \). Since each \( S_i \) is sparse, for each length \( n \
Rollout 4
.003605\n\]\n\nNext, we consider the voltage \( V = 5
Rollout 4
008775\n\]\n\nConverting this to a percentage:\n\n\[\n\text{
Top 16 Negative Activations
Rollout 8
the number of ordered sequences is 10P5. If they are indistinct, then the
Rollout 8
5 cards in order would be 10P5, as earlier.\n\nBut the problem here is that
Rollout 8
0 different cards, hence the number of ordered sequences is 10P5. If they are ind
Rollout 8
7 * 6 = 30240. However, in the color sequence RRRRR
Rollout 0
assigned to a or b, so 2^3=8 coprime pairs. Then half of
Rollout 8
10P5 = 30240. However, each color sequence (like RRGG
Rollout 7
Thus N=36 works. So possible N are 12,24,36.
Rollout 0
be 2^8 /2 =128. So 128 coprime pairs
Rollout 8
number of possible sequences is 30240.\n\nNow, we need to count how many of
Rollout 8
to our method:\n\nTotal sequences: 4P2=12\n\nHappy sequences: All red (
Rollout 8
replacement and the order matters. Alternatively, the number is 10P5 = 10!
Rollout 0
ime assignments (with order) is 2^3=8, but how many of those assignments result
Rollout 0
pairs (a,b) would be 2^8 = 256. However, because
Rollout 8
where she lays out 5 cards in order would be 10P5, as earlier.\n\nBut
Rollout 5
1. Therefore, the answer is 181. \n\nWait, but let me verify once again
Rollout 2
the remainder upon division by 1000 is109. But wait, but here's

DOWN_PROJ

Top 16 Positive Activations
Rollout 7
42)=14N/3\n\nMultiply both sides:\n\n(A -42)= (14
Rollout 8
5 red cards, but since they are identical? Wait, wait, no. Wait, the problem says
Rollout 1
other intersection theorem, which states that if we have a family of closed convex sets where every finite intersection is
Rollout 5
+ x) = 125\n\nMultiply both sides by (75 + x):\n\n(x +
Rollout 2
9 mod1000.\n\nWait, but hold on, earlier we got x=125
Rollout 7
) = 2N/3\n\nDivide both sides by N (assuming N ≠0, which
Rollout 6
the input i is between 1 and k.\n\nWait, but how does S know about k? Because
Rollout 7
42*(N/12)\n\nDivide both sides byN:\n\nA/12=1
Rollout 5
)] = 125H\n\nDivide both sides by H (assuming H ≠ 0):\n\n
Rollout 6
3 would give p(n) = 4 and 9, different. So if p is injective
Rollout 0
/ 2 = 128. But hold on, is this correct?\n\nWait, perhaps not
Rollout 1
For two spheres, it's a sphere of one dimension less.\n\nFor three points, x1, x
Rollout 7
4 -A)*(N/12)\n\nMultiply both sides by12:\n\n-24N=
Rollout 4
use the partial derivatives method. The formula for the uncertainty ΔC_eq would be sqrt( (��C
Rollout 1
and in a Hilbert space, the intersection of any number of closed convex sets is closed and convex.
Rollout 3
(π * M))\n\nR = 8.314 J/(mol·K)\n\nAssume
Top 16 Negative Activations
Rollout 8
4/3024=744÷24=31; 302
Rollout 8
= Let's check by dividing numerator and denominator by 240: 7440÷
Rollout 1
/ 2,\n\nwhich is ⟨x, x⟩ - 2⟨x, y⟩ + ⟨
Rollout 3
textbooks usually use 298 K (unless stated otherwise), perhaps answer around \boxed{16
Rollout 3
is the molar mass. Then there's the average speed, which is sqrt(8RT/(π
Rollout 6
. If yes, then x was in S_i, otherwise not. But is the encoding invertible?
Rollout 6
) consists of all tuples \((i, x)\) where \( x \in S_i \)
Rollout 1
Assume H has an orthonormal basis. If we have such a y, then the vectors { sqrt
Rollout 1
Hilbert space, given two vectors u and v, the parallelogram law says that ||u +
Rollout 8
all red or all green sequences are also acceptable, like RRRRR or GGGGG.\n\nSo in
Rollout 4
597*24=14,328 and 597*0.
Rollout 0
, others to b: a=5, b= 144 (16*9).
Rollout 0
as a fraction \(\frac{a}{b}\) where \(0 < a < b\) and
Rollout 8
\times 5 \times 4) = 2400\)\n - \(k
Rollout 7
score after increase is71*24=1704. Okay, so totals adds up
Rollout 6
is a perfect kth power, there's n, otherwise not. So for S_L = { α

Layer 56

GATE_PROJ

Top 16 Positive Activations
Rollout 0
=2^{8-1}=128.\n\n**Final Answer**\n\boxed{128
Rollout 2
09.\n\nTherefore, answer:109.\n\nBut let me recheck all steps once again to
Rollout 8
m=31, n=126.\n\nWait, but let's confirm. Wait, when
Rollout 8
1 +126= 157.\n\n**Final Answer**\n\boxed{157
Rollout 2
=1000n +109, meaning x≡109 mod100
Rollout 2
product is divisible by 1000, so the remainder is 0. However, given that
Rollout 8
31, n=126, which are coprime, so sum m +n=
Rollout 8
126 are coprime, so yeah, simplified.\n\nTherefore, the probability is 31
Rollout 4
the percentage error is approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation
Rollout 2
same as k=3, -1 mod8\n\nAll terms from k=3 onward are -1
Rollout 5
, 181 is the correct answer. \n\n**Final Answer**\n\boxed{181
Rollout 2
\n\n2^4 = 0 mod 8\n\nWait, 2^3 = 8 �
Rollout 5
.25 is 181. Therefore, final answer is 181. \n\nI
Rollout 8
and 126 are coprime, so yeah, simplified.\n\nTherefore, the probability is
Rollout 9
.\n\nTherefore, the altitude h is25. So the answer is 25. Therefore box form
Rollout 0
, hence The answer would be 128.\n\nBut before closing, verifying with our 6
Top 16 Negative Activations
Rollout 1
2)} / 2.\n\nBut perhaps it's hard.\n\nWait, back to the equation:\n\nFor y
Rollout 1
orthocentric system or something similar?\n\nLet me think.\n\nSuppose we fix y. Then the conditions
Rollout 1
a_i such that y exists.\n\nBut seems difficult to see.\n\nAlternative idea. From the properties we earlier
Rollout 9
to EM, h can be anything. Let's think. If h is less than 10,
Rollout 1
coefficients a_i such that y exists.\n\nBut seems difficult to see.\n\nAlternative idea. From the properties we
Rollout 1
be uncountable. Hmm.\n\nBut let's think: Suppose that S is a set where all pairwise
Rollout 9
,0).\n\nWait, wait. Let's step back. The triangle GEM is outside except for the
Rollout 6
, but the set must be sparse.\n\nLet's think. For each n, let’s denote a unique
Rollout 6
possible extensions. But that's exponential.\n\nThis seems not helpful. Let's reorient.\n\nPerhaps apply the
Rollout 2
1*11.\n\nBut maybe this isn't helpful. The key idea is to compute the P modulo
Rollout 8
are of two colors, perhaps?\n\nWait, maybe not. If each card is unique (distinct), even
Rollout 9
distance from G to EM.\n\nHmm. Let me try to imagine this. Since point G is the apex
Rollout 1
the same Hilbert space.\n\nWait, let me think again.\n\nGiven set S wherein between any two points
Rollout 1
- ) ) / 2.\n\nBut not making progress.\n\nWait but if subtract the two equations for x
Rollout 9
Wait, perhaps my assumption is incorrect. Let's recast this. I need to check whether when h
Rollout 1
can have arbitrary cardinality. However, let me refocus.\n\nSuppose that I can write each x

UP_PROJ

Top 16 Positive Activations
Rollout 0
2,3,5,7 are even?\n\nWait floor operations:\n\nFor prime 2:
Rollout 2
999*10 +9, wait, perhaps 9999=9*1
Rollout 5
(725). Is that right?\n\nWait, let me check the calculations again step by step.
Rollout 4
5000^2) )^2 )\n\nWait, this is another way of writing the terms
Rollout 4
) * 15 )^2\n\nHmm, let's compute these step by step.\n\nFirst,
Rollout 3
416*0.222) )\n\nFirst compute numerator:\n\n8*8.31
Rollout 5
* sqrt(725). Is that right?\n\nWait, let me check the calculations again step by
Rollout 8
2: 2400.\n\nWait, but hold on. Let me check for k=
Rollout 2
-1 for anyk is always even? No. When k=1: 9 is odd.
Rollout 2
0^k is 0 mod125?\n\nWait, 10^1 =10
Rollout 3
π * 0.222)) Hmm, okay. Let me calculate.\n\nFirst, 8
Rollout 0
2,3,5,7 are even?\n\nWait floor operations:\n\nFor prime 2: 2
Rollout 2
1=101*11? Wait: 1111=101*
Rollout 0
,5,7 are even?\n\nWait floor operations:\n\nFor prime 2: 20/2
Rollout 2
k is 0 mod125?\n\nWait, 10^1 =10 mod1
Rollout 2
1.\n\nSo first compute 9*99.\n\n9*99=891.
Top 16 Negative Activations
Rollout 1
are non-empty. So perhaps we need the family to be "centered" (has the finite intersection
Rollout 0
.\n\nAnd this number is equal to 2^{k -1}, where k is the number of distinct
Rollout 1
in particular, the set S would be in a one-to-one correspondence with an orthonormal system.\n\n
Rollout 9
\right) \, dx = 100 - \frac{500}{h}\n
Rollout 4
C2))\n\n= dC1 [1/C1 - 1/(C1 + C2)]
Rollout 6
most p(n) strings of length n. Since there are k sparse sets, maybe we combine them into
Rollout 1
{d}{\sqrt{2}} \) can be non-empty due to the finite intersection property and
Rollout 1
) must be orthogonal. This implies \( \langle x - y, x' - y \rangle =
Rollout 1
after scaling.\n\nWait, let me recall that if you have a regular simplex in R^n with edge length
Rollout 1
sqrt(2))2 ) = sqrt( d2/2 + d2/2 } ) =
Rollout 1
, this simplifies to:\n\nd2 = d2 - 2⟨x - y, x'
Rollout 6
most p(n) strings per length n. If k is a constant, then the total number for each
Rollout 4
respect to a is (b*(a + b) - a*b)/(a + b)^2)
Rollout 1
L(x) = ⟨x, 2y⟩ being equal to ⟨x, x⟩ + k
Rollout 5
that line is (a + x)/2 * k = 1/2 * [(a + b
Rollout 9
{10}{h}x + 10 - \frac{100}{h}

DOWN_PROJ

Top 16 Positive Activations
Rollout 6
with a sparse oracle. (This is the Meyer, I think, theorem.) So perhaps this is the
Rollout 4
percentage error is approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of
Rollout 8
+126= 157.\n\n**Final Answer**\n\boxed{157}\n\n
Rollout 0
2^{8-1}=128.\n\n**Final Answer**\n\boxed{128}\n\n
Rollout 3
\boxed{169} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon
Rollout 5
181 is the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n
Rollout 1
exists as described}}" or as per problem statement.\n\n**[Final Answer]**\n\boxed{\text{
Rollout 9
the altitude being h=25.\n\nPerfect.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>
Rollout 6
polynomial-time machine \( M \). By the Meyer's theorem equivalence between \( \textbf{P}/
Rollout 6
machine with a sparse oracle. (This is the Meyer, I think, theorem.) So perhaps this is
Rollout 4
and twice the relative error in V (because V is squared). But first, I need to find the
Rollout 6
a polynomial-time machine \( M \). By the Meyer's theorem equivalence between \( \textbf{P
Rollout 9
assistant.\n<|im_start|>user\nSquare $AIME$ has sides of length $10$ units.
Rollout 6
α_n for each input length n, similar to P/poly, but here the advice is fixed for
Rollout 3
requires two sig figs. 222 is three, so answer should keep three. So \
Rollout 1
in H. If S is a set of points each lying on the sphere of radius d/sqrt(
Top 16 Negative Activations
Rollout 4
)^2\n\nHmm, let's compute these step by step.\n\nFirst, compute the coefficients for Δ
Rollout 4
=5000\n\nCompute b2/(a + b)^2 = (3000
Rollout 4
.87%.\n\nBut wait, let's go step by step for verification once more.\n\nFirst, Compute
Rollout 4
Calculating the first term: (b2/(a + b)^2 * Δa)^2 =
Rollout 6
length. But maybe if we fixed the encoding to use the same number of bits for each n. Wait
Rollout 4
_eq = sqrt [ ( (b2 / (a + b)^2 ) * Δa )^
Rollout 3
π*M))\n\nLet's calculate the numerator and denominator step by step.\n\n8*R*T = 8 *
Rollout 7
there must be a mistake here. Let me verify step by step.\n\nWe derived:\n\nA*x +4
Rollout 4
.6 pF\n\nSecond term a2/(a +b)^2= (2000
Rollout 4
C_eq = sqrt( ( (b2)/(a + b)^2 * Δa )2 +
Rollout 4
, but because (C1*C2)/(C1 + C2), if both C1 and C
Rollout 4
)^2= 0.0000130000...\n\n(0.0
Rollout 4
C_eq with respect to a is (b*(a + b) - a*b)/(a + b
Rollout 5
5/(75 + x) = [(75 + x) - 125]/(
Rollout 6
in poly-time given |x|, which is the length of x, then the combined machine would first
Rollout 4
b*(a + b) - a*b)/(a + b)^2) = (b(a +

Layer 57

GATE_PROJ

Top 16 Positive Activations
Rollout 8
,k) * P(5,5 -k)]\n\nWait, for k=1:\n\nNumber for each
Rollout 4
2000 * 5000) = (30,000)/(1
Rollout 5
100))/2 * (h/2) = (2b + 150)/
Rollout 9
would be inside. If G is inside the square.\n\nWait, but earlier when h=10,
Rollout 4
dC2 / (C1 + C2))\n\n= dC1 [1/C1 -
Rollout 0
then we need to divide the total by 2.\n\nWait, but the problem says "how many rational
Rollout 9
area formula:\n\nAverage of the two bases * height.\n\nBut the bases are the two vertical sides. Wait
Rollout 4
, ideally if C1 and C2 are maximum.\n\nWait, but because (C1*C2)/(
Rollout 6
from n, which is polynomial in |x|.\n\nSo if the advice can be generated in polynomial time
Rollout 5
x = 5*sqrt(725)\n\nTherefore, x squared is 25*7
Rollout 0
of coprime ordered pairs, or is it?\n\nWait, but perhaps the counting is as follows:
Rollout 5
125]/(75 + x) = (x - 50)/(75
Rollout 8
green cards to the green positions. Let me see.\n\nFor a specific color sequence, say RRGGG
Rollout 7
42*( (N/3) -x).\n\nSimplify:\n\nA*x=56N/
Rollout 7
-5. Therefore, original average=C -5.\n\nWait, this approach seems different. Let me formal
Rollout 8
1))=2*(2*2)=8.\n\nWait, but according to our general formula:\n\nCase
Top 16 Negative Activations
Rollout 1
$ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show
Rollout 3
radon (Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to
Rollout 7
of non-promoted $47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm
Rollout 6
$L$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work
Rollout 7
-promoted $47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to
Rollout 2
$1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here
Rollout 3
Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean
Rollout 7
oted $47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve
Rollout 0
{}!$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure
Rollout 2
by $1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem
Rollout 9
$EM$ in $\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need
Rollout 6
language $L$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try
Rollout 2
is divided by $1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have
Rollout 8
. Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We
Rollout 6
L$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through
Rollout 0
{}^{}!$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need

UP_PROJ

Top 16 Positive Activations
Rollout 9
to make this clearer.\n\nLet's place the square on a coordinate system. Let's say point A is
Rollout 9
on the left side of EM. Wait, in our coordinate system, EM is at x=10
Rollout 9
make this clearer.\n\nLet's place the square on a coordinate system. Let's say point A is at
Rollout 2
? mod 1000\n\nSolve this system using Chinese Remainder Theorem.\n\nWe need
Rollout 8
of possible sequences would be different. But I think in probability problems with colored cards, unless specified, we
Rollout 9
. Wait squares have four sides. Let me actually assign coordinates to the square to make this clearer.\n\nLet
Rollout 0
128. But hold on, is this correct?\n\nWait, perhaps not quite. Since not
Rollout 2
the P modulo1000. But from the problem, rather than factorizing each term, might
Rollout 9
the base of the triangle is EM, which in this case is the vertical side. However, an is
Rollout 9
So that suggests something's wrong.\n\nWait, perhaps my assumption is incorrect. Let's recast this.
Rollout 2
5. Wait, in this case,\n\nbut solving the system leads x=1000n +
Rollout 8
as the permutations of the actual cards.\n\nBut maybe in terms of generating the happy sequences:\n\n- For each
Rollout 7
the increase.\n\nTherefore, to handle this, we need to model the original groups and the new groups.\n\n
Rollout 8
of permutations of the 5 cards, but actually this is confusing.\n\nWait, maybe not. Let's
Rollout 2
because the numbers are huge. So, perhaps there's a pattern or periodic behavior we can exploit when multiplying
Rollout 5
squared is involved? Hmm, maybe I need to use similar figures. If the trapezoid is
Top 16 Negative Activations
Rollout 3
^2 = 28,224. 169^2 = 28
Rollout 3
root.\n\nsqrt(28,436). Let's see, 168^2
Rollout 2
k>=3 are10^k -1. For10^3=1000
Rollout 3
^2 = 28,561. So sqrt(28,436)
Rollout 4
001) is 0.01. Let's approximate:\n\n0.0000
Rollout 7
2=14/3 +3.5=14/3 +7/2= (
Rollout 7
. However, the problem states N<40. 36 is less than 40,
Rollout 4
000 = 0.005. And for voltage V, ΔV / V =
Rollout 1
2} approx 0.707d. So sum of radii l_s = d /
Rollout 2
For k>=3:2^k mod8. But 2^3=8≡0 mod
Rollout 8
a block of 1 red and 1 green. However, actually each RG and GR in this case
Rollout 0
is counted twice except when \(a = b\). But since \(a \times b = 2
Rollout 3
Alternatively, sqrt(28,408). Let's see.\n\n168^2 =
Rollout 2
≡5 mod8 (109 mod8=5), so conclusion is x=109
Rollout 2
1: 891mod125. 125*7=875
Rollout 0
*9=144, b=5. Now a=144 and b=5

DOWN_PROJ

Top 16 Positive Activations
Rollout 5
areas, and then find the greatest integer not exceeding x2/100. \n\nFirst, let
Rollout 4
, since the energy is proportional to C_eq * V2, the relative error in energy will be a
Rollout 1
_x, v_{x'}⟩ = (2/d2)⟨x - y, x' -
Rollout 3
73 K as STP, but I'm not sure. Alternatively, in the context of an example
Rollout 3
divided by (kg/mol) is (m2/s2). Square root is m/s. So unit
Rollout 4
says "percentage error", and depending on how precise they want, maybe two significant figures? 0.
Rollout 1
would be sqrt( (d / sqrt(2))2 + (d / sqrt(2))2
Rollout 4
%.\n\nBut the options or the answer expected? Possibly they want an answer to two decimal places. For example
Rollout 1
∈ S, ||x - y||2 = d2 / 2.\n\nExpanding, that gives
Rollout 1
If we have ||x - y||2 = d2 / 2 for all x, then from
Rollout 5
a + x)/2 * k = 1/2 * [(a + b)/2 * h
Rollout 3
which is same as (kg·m2)/(s2·mol·K). Then, 8
Rollout 4
energy is (1/2)*C_eq*V2, maximum energy would be when C_eq is
Rollout 3
0 /0.08796≈225,298. Then sqrt(
Rollout 3
798) ) �� sqrt( ~225,753 ) ��
Rollout 4
, because the energy is proportional to C_eq and V squared. Therefore, exponents are 1 for
Top 16 Negative Activations
Rollout 7
average54, total=54*9=486. Total=186+
Rollout 9
(10,10), which is the vertical side on the right. But wait, the problem
Rollout 8
31, n=126, which are coprime, so sum m +n=
Rollout 3
.314*298= about 19,820 (same as earlier
Rollout 8
1: \(2 \times 5! = 2 \times 120 = 2
Rollout 5
= 125y\n\nBring all terms to left side:\n\ny^2 - 25
Rollout 9
0,10), which is the vertical side on the right. But wait, the problem says the
Rollout 0
b\) and \(a\) and \(b\) are coprime positive integers. The product \(a
Rollout 0
they form a coprime pair whose product is 20!. Such pairs are related to the div
Rollout 3
8.314*300= 19,953.6\n\nDiv
Rollout 8
1: 2 * 5! = 2 * 120 = 24
Rollout 3
the average speed of N2 molecules at 25°C. And they do so using sqrt(8
Rollout 9
M is top-right, so that would be the vertical side. Wait, no, in that labeling,
Rollout 4
9*3.1623≈8.77.\n\nTherefore, 8.77
Rollout 9
GEM extends from the base EM (on the right edge of the square) towards point G which is
Rollout 1
... Wait, perhaps thinking of yi's dual variables.\n\nAlternatively, consider H as some sequence space, and

Layer 58

GATE_PROJ

Top 16 Positive Activations
Rollout 6
Meyer's theorem equivalence between \( \textbf{P}/\text{poly} \) and sparse
Rollout 3
0.222 �� 3.1415926536
Rollout 3
0.222 �� 3.1416 * 0.22
Rollout 6
P/poly is established that a language is in P/poly if and only if it is polynomial-time
Rollout 3
8)) �� sqrt( (8*8.314*298)/(3.
Rollout 6
string α_n for each input length n, similar to P/poly, but here the advice is fixed
Rollout 6
a sparse set $S$ and a deterministic polynomial time TM $M$ with oracle
Rollout 6
, the standard result is that a language is in P/poly if and only if it is Turing-re
Rollout 3
512 * 270 = 17,958.24, plus
Rollout 8
2 \times 5! = 2 \times 120 = 240\
Rollout 6
bad-angel}} \) does not provide the nondeterminism inherent in \( \textbf{NP
Rollout 3
*8.314 = 66.512\n\nThen, 66.
Rollout 6
set $S$ and a deterministic polynomial time TM $M$ with oracle access to
Rollout 3
8.314 = 66.512\n\nThen 66.5
Rollout 3
0.222 �� 0.697433\n\nThus, 1
Rollout 6
exists a polynomial $p : \mathbb{N} \mapsto \mathbb{N
Top 16 Negative Activations
Rollout 1
would be sqrt( (d / sqrt(2))2 + (d / sqrt(2))2
Rollout 5
areas, and then find the greatest integer not exceeding x2/100. \n\nFirst, let
Rollout 9
+ (0 - y)^2] = sqrt[h^2 + y^2]\n\nSimilarly, GM
Rollout 2
109≡5 mod8.\n\nTherefore, equation becomes:\n\n(5m +5)≡5
Rollout 1
parallelogram law says that ||u + v||2 + ||u - v||2 =
Rollout 9
the square is labeled AIME, moving around the square, then maybe A is the bottom-left, I
Rollout 5
8125 = 0\n\nWait, that leads to x2 = 1812
Rollout 2
0^k - 1).\n\nSo the product we are dealing with is P = (10^
Rollout 4
so 0.88%, or maybe the answer is exact at 0.87%.\n\nBut
Rollout 1
x, y⟩ + ||y||2 = d2 / 2.\n\nSimilarly for another x',
Rollout 8
8 * 7 * 6. Alternatively, that's 30240. But let
Rollout 1
, x⟩ + ⟨y, y⟩ - d2 / 2)/2.\n\nHmm, but
Rollout 9
from G to EM. So h is the length we need to find.\n\nNow, the common area between
Rollout 4
F.\n\nTotal Uncertainty: sqrt(3.62 +2.42)=sqrt(1
Rollout 1
, setting c = ⟨y, y⟩ - d2 / 2.\n\nWait, from the equation
Rollout 5
MS) of the two bases. So sqrt[(a2 + b2)/2]. Let me check

UP_PROJ

Top 16 Positive Activations
Rollout 7
56N -56P=56N +15P=66N\n\nSo
Rollout 2
999=999*10 +9, wait, perhaps 9999
Rollout 7
9P +79x\n\n→-3P +A x -79x=0\n\n
Rollout 2
1) \equiv -891 \equiv -16 \equiv 109 \mod
Rollout 7
/3\n\nThus:\n\nx(A -42) +14N=56N/3
Rollout 7
6N -56P=56N +15P=66N\n\nSo
Rollout 7
6N -56P = 56N +15P=66N\n\nSo
Rollout 7
part(a):\n\nA*x +42*N/3 -42x=56N/3
Rollout 7
213N -205N)/3=8N/3.\n\nHence:\n\n3
Rollout 7
ide by N (N≠0):\n\n-2 + (A -79)/12=0
Rollout 9
(-5/h)x + (50/h) +0? Wait, let's verify:\n\nFrom G
Rollout 9
to10] [ (1/5 x +8 +1/5 x -2 ) ]
Rollout 4
328 and 597*0.8004≈ 597*
Rollout 7
Divide by N (N≠0):\n\n-2 + (A -79)/12=
Rollout 7
*(N/12)=14N/3 +42*(N/12)\n\nDiv
Rollout 7
N/12)=14N/3 +42*(N/12)\n\nDivide
Top 16 Negative Activations
Rollout 2
satisfies the first congruence.\n\nWait, hmm, that seems contradictory. Wait, no, according to
Rollout 2
8 is5 which is correct. Hmm. Wait, am I missing something here?\n\nWait, in our
Rollout 2
, which satisfies the first congruence.\n\nWait, hmm, that seems contradictory. Wait, no,
Rollout 9
, the horizontal distance from G to EM.\n\nWait, maybe visualizing coordinates would help. Let's place
Rollout 4
But error propagation using partial derivatives is more efficient here.\n\nHowever, let me verify to make sure.\n\nThe
Rollout 4
4%), so maybe a couple of decimal points. \n\nAlternatively, perhaps compute the maximum and minimum possible values
Rollout 0
20! but 7^3 does not.\n\nBut exponent of 7 is 2, which
Rollout 9
0,0) lies within the square? Yes.\n\nWait, but then the area of the triangle in
Rollout 9
, meaning legs GE and GM are equal. Wait, if it's isosceles with base EM
Rollout 5
Therefore, the answer is 181. \n\nWait, but let me verify once again. The
Rollout 2
9 mod8 is5 which is correct. Hmm. Wait, am I missing something here?\n\nWait,
Rollout 5
ratio 2:3. \n\nWait a sec, the midline divides the trapezoid into
Rollout 2
. Hmm. Wait, am I missing something here?\n\nWait, in our calculation, after solving we had
Rollout 9
so the triangle is completely within the square. Wait, because the square spans from x=0 to x
Rollout 8
0.\n\nThus, I think the calculation is accurate.\n\nAn alternative validation: Let's compute the total number
Rollout 5
not exceeding this is 181.\n\nWait, this seems straightforward now, but maybe I had a

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
98) ) �� sqrt( ~225,753 ) �� 4
Rollout 3
/0.697433≈26,034.21.\n\nsqrt
Rollout 3
/ 0.697 �� 26,049.\n\nsqrt(26
Rollout 3
/0.697433≈26,050\n\nsqrt(26
Rollout 1
where all the distances are equal, it's called an equidistant set. In Hilbert spaces,
Rollout 3
/0.08796≈225,298. Then sqrt(2
Rollout 1
space is compact. However, Hilbert spaces are only locally compact in finite dimensions, thus only in finite
Rollout 6
think I'm going in circles here. Given the time constraints, maybe for problem 3, since it
Rollout 1
-separable Hilbert spaces, you have uncountable orthonormal bases, while in separable only
Rollout 7
repeaters, N = P + R.\n\nWe are given:\n\n1. The average of all participants is
Rollout 1
bert spaces, you have uncountable orthonormal bases, while in separable only countable ones
Rollout 1
typically in non-separable Hilbert spaces, you have uncountable orthonormal bases, while in
Rollout 1
the dimension is the cardinality of any orthonormal basis. But H here is infinite-dimensional, which
Rollout 3
406, then sqrt≈168.5 m/s. So insignificant change.\n\nAlternatively,
Rollout 2
125*(8n)+109=1000n +109\n\n
Rollout 1
something called a "pseudobasis" or "orthonormal basis" even if uncountable
Top 16 Negative Activations
Rollout 3
but as per my knowledge cutoff is 2023, so if need to answer, in the
Rollout 4
00 /4975= 4. Therefore, C_eq_min=1190
Rollout 3
2 * 270 = 17,958.24, plus 6
Rollout 7
leading to a max average of64.\n\nThis is impossible, meaning that assuming the given conditions(after increase
Rollout 7
scored 64). So A=98 is impossible.\n\nTherefore, inconsistency. What does that mean
Rollout 7
2 +56=98.\n\nBut that's impossible as A must be <=64 (since
Rollout 8
the lengths of each block are variable (must be at least 1, obviously). But the key is
Rollout 8
, 31/126≈0.246.\n\nTo verify if that's correct
Rollout 4
00 + 3000) = (6,000,000)/
Rollout 4
sqrt(7.7) is approx 2.7749, so sqrt(7.
Rollout 4
(10^-5) = 10^-2.5 �� 3.16
Rollout 3
19,820.576.\n\nDivide that by π * 0.2
Rollout 3
0.222 ��0.697433.\n\n19,8
Rollout 4
+ 3000} = \frac{6000000}{50
Rollout 1
r2 = d sqrt(2) �� 1.414d >= d, so
Rollout 4
2000*3000 = 6,000,000;

Layer 59

GATE_PROJ

Top 16 Positive Activations
Rollout 7
66N +5N =71N.\n\nBut separately, we can say total promoted score
Rollout 7
59(R - x) = 71N.\n\nAdditionally, we have original relations:\n\nOriginal total
Rollout 7
) = 76P + 61R = 66N +5N =7
Rollout 5
50))/2 * (h/2) = (2b + 50)/2
Rollout 7
for part(b) is N=24?\n\nWait, wait. Earlier we saw that in equations,
Rollout 1
c - d2/2) ) = 0 ⇒ (2c - d2)/2 -
Rollout 0
pairs (a,b) would be 2^8 = 256. However, because
Rollout 1
parallelogram law says that ||u + v||2 + ||u - v||2 =
Rollout 2
=999≡-1 mod125.\n\nIn fact, as noticed before, 1
Rollout 7
participants would be 71P + 5P = 76P, and for repeaters
Rollout 8
]\n\nSo total for Case 2:\n\nSum from k=1 to 4 of [2 * (
Rollout 1
x, y⟩ + ||y||2 = d2 / 2.\n\nSimilarly for another x',
Rollout 3
some textbooks might use mean as rms. It's unclear.\n\nAlternatively, to check. Maybe the mistake is
Rollout 7
.\n\nOriginal total repeater score: 56R = 56(N - P).\n\nSo total
Rollout 7
from original promoted:71 +5 =76. Wait, original promoted had average 71
Rollout 7
total score becomes 66N + 5N = 71N. But also, the
Top 16 Negative Activations
Rollout 1
convex. But since H is infinite-dimensional, it is a complete metric space. However, the intersection might
Rollout 1
\mathcal{H}$ be an infinite-dimensional Hilbert space, let $ d>0
Rollout 3
(298 K), average speed is 1155 km/h, which translates to ~
Rollout 3
°C (298 K), average speed is 1155 km/h, which translates to
Rollout 1
, hence the desired y exists.\n\nSince H is a complete metric space, and the family of spheres has
Rollout 1
thonormal system must be countable because the space is separable? Wait, but the problem states that
Rollout 3
.512.\n\nThen, 66.512 * 298 = let
Rollout 1
-empty, hence the desired y exists.\n\nSince H is a complete metric space, and the family of spheres
Rollout 1
point \( y \) is guaranteed by the properties of infinite-dimensional Hilbert spaces, ensuring the required or
Rollout 1
the finite intersection property is non-empty if the space is compact. However, Hilbert spaces are only locally
Rollout 1
the whole intersection is non-empty, provided the space is complete (which it is) and perhaps the family
Rollout 1
. But since H is infinite-dimensional, it is a complete metric space. However, the intersection might still
Rollout 1
a non-empty intersection. In infinite dimensions, balls are not compact hence there is no guarantee.\n\nHences
Rollout 1
(has the finite intersection property) and maybe in a weakly compact set. But in Hilbert spaces
Rollout 1
the space is compact. However, Hilbert spaces are only locally compact in finite dimensions, thus only in
Rollout 1
that in an infinite-dimensional Hilbert space, even a countable set can be dense.\n\nAlternatively, we

UP_PROJ

Top 16 Positive Activations
Rollout 1
x'.\n\nLet me recall that in Hilbert spaces, if we have a set of vectors with equal norms
Rollout 1
for distinct x, x'.\n\nLet me recall that in Hilbert spaces, if we have a set of
Rollout 0
in 20!:\n\nFor prime 2: floor(20/2) + floor(
Rollout 8
4/3024=744÷24=31; 302
Rollout 4
72). sqrt(18.72)= approximately 4.326 pF.\n\n
Rollout 1
ormal systems can be uncountable. However, in that case, the problem still requires that {v
Rollout 9
vertical, then the base is vertical. But triangles are usually thought of as having a horizontal base, but
Rollout 1
of a simplex or something similar. For example, in finite dimensions, if you have points on a sphere
Rollout 8
126 divides by 2, 3, etc. 126 is 2*
Rollout 5
, the midline divides the trapezoid into two smaller trapezoids? Each with height
Rollout 1
zero for distinct x, x'.\n\nLet me recall that in Hilbert spaces, if we have a set
Rollout 1
where each pair is orthonormal. However, in Hilbert spaces, the dimension is the cardinality
Rollout 1
i.e., tetrahedron, but even a triangle is 2D.\n\nWait, maybe in
Rollout 4
4975*1190=4975*(1200-
Rollout 3
8.5 is exactly halfway. Rounding rules: round to the nearest even, so 16
Rollout 1
vertices of a simplex or something similar. For example, in finite dimensions, if you have points on a
Top 16 Negative Activations
Rollout 6
with α_n | n ∈ N }, then this set is sparse.\n\nTo retrieve α_n, the oracle
Rollout 4
series. The voltage applied across this combination is $V=5.00 \pm 0.
Rollout 1
distance from 0 to an arbitrary z in S' is ||z|| = ||x - x0
Rollout 6
be the binary encoding of α_n. But α_n has length p(n), so s_n is a
Rollout 8
5,5−r).\n\nSo:\n\nTotal happy sequences = sum over r=0 to 5 [
Rollout 6
= { α_n | n ∈ N }, each string is of length p(n), hence the number of
Rollout 1
such that { x - y : x ∈ S'} are orthogonal.\n\nWait, original problem: when translating
Rollout 1
if r1 + r2 >= l and |r1 - r2| <= l.\n\nIn our
Rollout 4
. The relative error here is ΔC1 / C1 = 10 / 200
Rollout 4
term inside the formula is [ (C2 ΔC1 ) / (C1 (C1 +
Rollout 3
.\n\nsqrt(26,034.21)= approximately 161.35
Rollout 4
_eq/∂C1= [C2/(C1 + C2)]^2 = (3
Rollout 6
has length p(n), so s_n is a string of length p(n). Thus, S_L =
Rollout 6
of length $n$ in $S$ is bounded by $p(n)$
Rollout 4
C2)\n\nThen combine terms:\n\n= (dC1 / C1 - dC1 / (
Rollout 2
solution is x≡109 mod1000.\n\nWait, but hold on, earlier we

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
terms of molar mass, since it's a monatomic gas, the molar mass would be approximately
Rollout 6
is similar to problem 1. If you consider countably infinite advice strings for all n, and represent
Rollout 7
71N -205N/3=(213N -205N)/
Rollout 0
.\n\nTherefore, the previous answer 128 must be correct, hence The answer would be 1
Rollout 1
, ||x - y|| = d / sqrt(2) for all x in S. So every
Rollout 1
the vector x - y and x' - y must be orthogonal. That suggests that the inner product is
Rollout 7
/3 +7/2= (28 +21)/6=49/6≈
Rollout 7
)/12=2→A-79=24→A=103.\n\nBut
Rollout 3
the isotope, then its molar mass is 222 g/mol, or 0.
Rollout 3
. At STP, the molar volume is 22.4 liters, which is for ideal
Rollout 7
24 works.\n\nN=36:\n\nP=24, R=12.\n\nx=
Rollout 1
. Then, for each x ∈ S \ {0}, we have:\n\n||x - y||2
Rollout 1
S lie on a sphere of radius d/sqrt(2) around y, and any two points x
Rollout 3
.4, as 161^2=25,921;162
Rollout 7
A)*N\n\nAssuming N ≠0:\n\n-24=74 -A\n\nTherefore,A
Rollout 7
After +5:1704+5*24=1704+12
Top 16 Negative Activations
Rollout 7
average54, total=54*9=486. Total=186+
Rollout 9
= sqrt[(10 - (10 - h))^2 + (0 - y)^2]
Rollout 9
:\n\nThe slope (m) is (0 -5)/(10 - (10 - h))
Rollout 3
8,436). Let's see, 168^2 = 28,
Rollout 4
2 being C22/(C1 + C2)^2 and C12/(C1 +
Rollout 1
Hilbert space is istit's isomorphic to l2(I) for some index set I. Therefore
Rollout 5
0\n\nThis is a quadratic in y: y2 -150y -125
Rollout 3
3,476.2) �� 183 m/s. So that's the
Rollout 5
areas, and then find the greatest integer not exceeding x2/100. \n\nFirst, let
Rollout 1
x, y⟩ + ||y||2 = d2 / 2.\n\nSimilarly for another x',
Rollout 1
.\n\nWe know ||x - x'||2 = d2, ||x - y||2 = (
Rollout 1
parallelogram law says that ||u + v||2 + ||u - v||2 =
Rollout 4
But the error here is calculated as (Max - Min)/(2*Nominal). Wait, (1
Rollout 9
0)):\n\nSlope is (10 -5)/(10 - (10 - h))
Rollout 1
translated have 〈x, x'〉 = d2 / 2, and we set ||y
Rollout 7
. Let's compute that.\n\nFirst expand:\n\n75*(2N/3) +75x

Layer 60

GATE_PROJ

Top 16 Positive Activations
Rollout 5
zoid into two regions whose areas are in the ratio $2: 3$ . Let $x
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capac
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the
Rollout 5
the trapezoid into two regions whose areas are in the ratio $2: 3$ .
Rollout 8
5$ red cards and $5$ green cards. She shuffles the $10$ cards and
Rollout 6
a deterministic polynomial time TM $M$ with oracle access to $S$ such that
Rollout 6
} \mapsto \mathbb{N}$ , a sequence of strings $\{\alpha_n\
Rollout 8
$ red cards and $5$ green cards. She shuffles the $10$ cards and lays
Top 16 Negative Activations
Rollout 1
= 2d / sqrt(2) = d sqrt(2) > d, so the spheres
Rollout 1
y||2. Therefore:\n\nd2 = (d2)/2 + (d2)/2 -
Rollout 0
a^2 < 20! or a < sqrt(20!). Therefore, the number of
Rollout 1
vectors each of length d, with pairwise inner product d2/2.\n\nThen, in this case,
Rollout 1
translated have 〈x, x'〉 = d2 / 2, and we set ||y
Rollout 1
x - y, x' - y) = d2 / 2 - 3d2/
Rollout 4
= sqrt( ( (b2)/(a + b)^2 * Δa )2 + ( (
Rollout 1
Therefore, this simplifies to:\n\nd2 = d2 - 2⟨x - y, x
Rollout 3
28,408.12:\n\nsqrt(28,408.1
Rollout 1
pair in S \ {0} has inner product d2/2. Then the set is a collection
Rollout 4
.4 pF\n\nThus, ΔC_eq =sqrt(3.62 +2.4
Rollout 1
previous calculation, ⟨x, x'⟩ = d2 / 2.\n\nTherefore, if y is
Rollout 1
/4 + d2 / 2 = (d2 / 2 + d2 / 2
Rollout 1
||y|| squared ) / 2 = (d2 + d2 / 2)/2 =
Rollout 1
uting values:\n\n〈x, x'〉 = d2 / 2,\n\n〈x, y〉
Rollout 4
)^2 * Δa )2 + ( (a2)/(a + b)^2 * Δb

UP_PROJ

Top 16 Positive Activations
Rollout 5
In a trapezoid, the midline (the segment connecting the midpoints of the legs)
Rollout 1
space, any orthonormal system must be countable because the space is separable? Wait, but
Rollout 1
points in R^3 with equal pairwise distances d (i.e., an equilateral triangle in 3
Rollout 9
(length 10 -0=10) and height=10 (the distance between x
Rollout 1
there is something called a "pseudobasis" or "orthonormal basis" even if un
Rollout 5
same. But in general, trapezoids are not similar unless the ratio of the bases is the
Rollout 0
be \(2^8 = 256\), but each pair is counted twice except when \(
Rollout 1
intersection property) and maybe in a weakly compact set. But in Hilbert spaces, closed balls are
Rollout 4
) - a*b)/(a + b)^2) = (b(a + b - a)) /
Rollout 1
honormal systems in Hilbert spaces must be countable if H is separable, but H is not
Rollout 1
approach: let's define y as the weak limit or something of the points in S, but in H
Rollout 1
perhaps using the concept of a orthocentric system or something similar?\n\nLet me think.\n\nSuppose we
Rollout 1
spaces, orthonormal systems can be uncountable. However, in that case, the problem still
Rollout 1
is something called a "pseudobasis" or "orthonormal basis" even if uncount
Rollout 1
order to have equidistant points, they must form a regular simplex, which can be embedded in n
Rollout 1
considering properties of such sets S. If S is a set where all the distances are equal, it's
Top 16 Negative Activations
Rollout 7
likely, the problem expects there to be solutions. So I must have made a mistake.\n\nWait, alternative
Rollout 7
. So I must have made a mistake.\n\nWait, alternative approach. Let me use another angle.\n\nGiven
Rollout 7
)\n - Substituting \( P = \frac{2N}{3} \) and \(
Rollout 7
given. But we have to check our steps.\n\nSo Part (b) asks: after 5 point
Rollout 2
=> 0 mod1000, so10^4 -1=9999
Rollout 7
is impossible.\n\nWait, that can't be. So there must be a mistake here. Let me verify
Rollout 3
mann distribution:\n\n\[\nv = \sqrt{\frac{8RT}{\pi M}}\n\]\n\n
Rollout 2
000=>-891=109 modmod1000. Hence,
Rollout 7
6 students average 54: total=54*6=324. Total=
Rollout 7
before. But how does this make sense?\n\nAlternatively, perhaps during the problem, the composition of promoted and
Rollout 4
F}$ and $C_{2}=3000 \pm 15 \mathrm{p
Rollout 2
=0 mod125, so 10^4 -1≡-1 mod12
Rollout 7
unless there are constraints I haven't considered.\n\nWait, perhaps non-integer averages? But average A=
Rollout 2
8\), so \(2^k - 1 \equiv -1 \mod 8\).\n
Rollout 4
[\nC_{\text{eq}} = \frac{2000 \cdot 30
Rollout 3
169 m/s.\n\nBut wait, perhaps my calculator computation precision was a bit too picky.

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
0.697 ��28,624; sqrt ��169
Rollout 4
327 / 12 = 0.360583333...
Rollout 3
/0.08796≈225,298. Then sqrt(2
Rollout 7
16 with total 71*16=1136. Increase by 5 each
Rollout 2
0 mod125\n\n10^3=1000 mod125.
Rollout 3
.697433≈28,622\n\nsqrt(28,6
Rollout 8
contiguous color blocks. Hence, such sequences are called "linear arrangements with two blocks".\n\nAlternatively, in comb
Rollout 7
x=2, sum=98*2=196. Original scores:avg=1
Rollout 3
98) ) �� sqrt( ~225,753 ) �� 4
Rollout 7
average62, total=62*3=186. Remaining 9 with average5
Rollout 8
order matters, the number of sequences is the number of injective functions from 5 positions to 1
Rollout 3
222, with a half-life of about 3.8 days. But in terms of m
Rollout 9
Wait, but in trapezoid area formula:\n\nAverage of the two bases * height.\n\nBut the
Rollout 4
^2) = (9,000,000)/(25,000
Rollout 3
8 +66.512*3=199.536 =19
Rollout 4
) = 9/25 = 0.36\n\nSo, 0.36
Top 16 Negative Activations
Rollout 1
balls are not compact hence there is no guarantee.\n\nHences perhaps requires another approach, using the structure of
Rollout 3
/0.697433≈27,950\n\nsqrt(27
Rollout 3
/0.697433≈27,950. sqrt(27
Rollout 3
016÷0.697≈27,950; sqrt(27
Rollout 3
75 /2.828≈168, which is about right. So, the
Rollout 3
512 * 270 = 17,958.24, plus
Rollout 5
areas, and then find the greatest integer not exceeding x2/100. \n\nFirst, let
Rollout 3
sqrt(27,950)=167.2 m/s.\n\nSo approximately 1
Rollout 3
/0.697433≈28,406, then sqrt≈1
Rollout 3
/ 0.697 �� 28,433. Let's take sqrt
Rollout 6
set. So that's probably the answer here.\n\nHence, you just reference this result. However,
Rollout 1
= d2 - (||x||2 + ||x'|||^2).\n\nDividing both sides by
Rollout 1
-2〈x, y〉 + ||y||2 = (d2) / 2 -
Rollout 3
sqrt(27,950)=167.2 m/s. So around 1
Rollout 0
are: 2, 3, 5, 7, 11, 13
Rollout 1
)/2.\n\nSo,\n\n(||x||2 + ||x'||2 - d2)/2 - ⟨

Layer 61

GATE_PROJ

Top 16 Positive Activations
Rollout 0
total pairs, so 256/2 =128. Hence 128 such
Rollout 0
a < b hence 256/2=128. So confirms.\n\nBut let me
Rollout 8
.\n\nThus, the answer would be 31 +126=157.\n\nBut I
Rollout 7
the equation becomes:\n\n209N/3 +16x =71N.\n\nNow,
Rollout 2
�5 mod8\n\nCompute 125 mod8: 125 /8=
Rollout 0
×3...×20 is given by 2^{number of prime factors of 20!
Rollout 2
5 mod8\n\nand P≡109 mod125.\n\nNow need to find a number
Rollout 7
satisfying given conditions.\n\nTherefore,(a) possibleN:12,24,36; (
Rollout 8
, so sum m +n= 31 +126= 157.\n\n**
Rollout 3
molecular speed" is average speed, sqrt(8RT/(πM)).\n\nTherefore, assuming room temperature
Rollout 7
\n\nAssuming N ≠0:\n\n-24=74 -A\n\nTherefore,A=74
Rollout 7
must be integer, so N must be a multiple of 12. Just like part (a).
Rollout 9
the square (to the left), at (10 - h,5) where h >10
Rollout 7
14N/3\n\nDivide both sides by N (N ≠0):\n\n(A -42
Rollout 9
2 + (0 - y)^2] = sqrt[h^2 + y^2]\n\nSimilarly,
Rollout 2
m≡0 mod(8/ gcd(5,8))=0 mod8. Therefore, m
Top 16 Negative Activations
Rollout 5
b + 25}{b + 75} = \frac{2}{3}\n\
Rollout 5
+ 25}{b + 75} = \frac{2}{3}\n\]\n
Rollout 9
0 - \frac{500}{h} = 80 \implies \frac{
Rollout 4
}{2000 + 3000} = \frac{60000
Rollout 9
00 - \frac{500}{h} = 80 \implies \frac
Rollout 8
Case 1: \(2 \times 5! = 2 \times 120 =
Rollout 5
frac{18125}{100} = 181.25\n
Rollout 5
{18125}{100} = 181.25\n\
Rollout 1
langle x - y, x' - y \rangle = 0 \).\n\n2. **Key Observ
Rollout 2
25\n \]\n - Let \(x = 125m + 10
Rollout 4
00000}{5000} = 1200 \, \text
Rollout 4
2000 + 3000} = \frac{600000
Rollout 0
\) is \(\frac{256}{2} = 128\).\n\nThus,
Rollout 5
25 + 30625}{2}} = \sqrt{\frac{362
Rollout 5
we find:\n\[\n3(b + 25) = 2(b + 75)
Rollout 0
airs**:\n - For \(a \times b = 20!\) and \(\gcd

UP_PROJ

Top 16 Positive Activations
Rollout 5
stituting these values:\n\[\nx = \sqrt{\frac{75^2 + 1
Rollout 5
175^2}{2}} = \sqrt{\frac{5625 + 3
Rollout 3
oltzmann distribution:\n\n\[\nv = \sqrt{\frac{8RT}{\pi M}}\n
Rollout 7
total:79*(P + x)+47*(R -x)=71N.\n\nSame
Rollout 5
30625}{2}} = \sqrt{\frac{36250}{2
Rollout 7
N/3) +79x +47*(N/3) -47x=
Rollout 8
: \(2 \times 5! = 2 \times 120 = 24
Rollout 0
1\), each prime factor of 20! must be assigned entirely to either \(a\)
Rollout 8
).\n - Total for Case 1: \(2 \times 5! = 2 \times
Rollout 0
in lowest terms is \(20!\) is \(\boxed{128}\).<|im_end|>
Rollout 7
9*(2N/3 +x)+47*(N/3 -x)=71N
Rollout 7
) are \(\boxed{12}\), \(\boxed{24}\), and \(\
Rollout 7
P +x), non-promoted total=47*(R -x). Then:\n\n79(P
Rollout 8
**\n - There are 2 such sequences: RRRRR and GGGGG.\n - Number
Rollout 5
) is given by:\n\[\nx = \sqrt{\frac{a^2 + b^2
Rollout 3
8 K.\n\nCompute v_avg = sqrt(8*R*T/(π*M))\n\nLet's calculate the numerator
Top 16 Negative Activations
Rollout 2
izing each term, might use modulus1000. However, since 1000=
Rollout 3
sqrt(8*R*T/(π*M))\n\nLet's calculate the numerator and denominator step by step.\n\n8
Rollout 2
. But wait when doing straight mod1000 I also arrived at109, which is
Rollout 3
mol unit cancels, then kg in denominator; thus overall (kg·m2)/(s2·
Rollout 2
5 is9 *99 * (-1). So, 9*99=891
Rollout 7
haven't considered.\n\nWait, perhaps non-integer averages? But average A=98 in original scores
Rollout 2
=3 is9*99*(-1). So from the previous step when we had9*9
Rollout 7
12, and A is the average, maybe that's okay.\n\nSo we have x = N/
Rollout 4
15 )^2\n\nHmm, let's compute these step by step.\n\nFirst, compute the
Rollout 2
09 mod1000.\n\nWait, so pattern for the progression:\n\nFirst term:9\n\n
Rollout 3
16*0.222) )\n\nFirst compute numerator:\n\n8*8.314
Rollout 4
.04%), so maybe a couple of decimal points. \n\nAlternatively, perhaps compute the maximum and minimum
Rollout 2
-1 mod1000. Therefore:\n\nSo after multiplying the first two terms, get 8
Rollout 2
=109 mod125. So that matches.\n\nThus, if we follow the steps,
Rollout 3
π * 0.222) )\n\nFirst calculate 8 * 8.314
Rollout 2
997 terms of-1 each. Hence, total product is891*(-1)^

DOWN_PROJ

Top 16 Positive Activations
Rollout 6
log n + n^k �� m? Unlikely. For example, m is dominated by n
Rollout 1
d2 ) / 2 = 0.\n\nRearranging:\n\n||y||2 - ⟨
Rollout 2
09 mod8: 109-13*8=109-10
Rollout 1
inner products.\n\nIf it's possible to find y such that translating by y makes the Gram matrix orthogonal,
Rollout 1
is d, I need to find a point y such that { sqrt(2)/d (x -
Rollout 2
is≡-1, so multiplying 997 terms of-1. However,-1^{9
Rollout 2
subsequent term (total of997 terms) is-1 mod1000. So multiplying
Rollout 3
.512 * 300 = 19,953.6 minus
Rollout 2
cdots \times \underbrace{99\cdots9}_{\text{999
Rollout 3
.512 * 300 = 19,953.6\n\nBut
Rollout 1
\left\{\frac{\sqrt{2}}{d}(x\minus{}y): \ x
Rollout 2
0\n\nMultiply by9999 (four nines):109*9999
Rollout 1
S, but in H, even an infinite set might not necessarily have any kind of limit. Maybe.\n\n
Rollout 3
root of the result.\n\n\[\n\text{Numerator: } 8 \times 8.
Rollout 1
to map the equidistant points into orthonornal through such a shift and scaling.\n\nSumming
Rollout 8
6 = 30240.\n\nNumerator:\n\nFirst, Case 1: 2
Top 16 Negative Activations
Rollout 0
20! or a < sqrt(20!). Therefore, the number of such a is half
Rollout 8
5 -k) = 5! / (k)!.\n\nSo the number for each color sequence (
Rollout 1
vectors in $ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 4
of the energy stored in this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 1
x' - y is zero for distinct x, x'.\n\nLet me recall that in Hilbert spaces,
Rollout 2
mod125.\n\nSince (-1)^{997}= -1.\n\nTherefore, 1
Rollout 4
008774964*100=0.87749
Rollout 5
2(b + 75); 3b + 75 = 2b + 1
Rollout 8
5 - k)!] * [5! / k!].\n\nThen, since for each k (from
Rollout 0
! are 2, 3, 5, 7, 11, 13
Rollout 3
sqrt(8*8.314*298/(π*0.028
Rollout 8
6. Alternatively, that's 30240. But let me verify that. Yes
Rollout 9
of 80? Since the square is 100, the triangle has to somehow overlap
Rollout 8
+ 1200) + (2400 +2400) =
Rollout 2
$ is divided by $1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I
Rollout 2
we have 16 * (-1)^{997} mod125.\n\nSince (-

Layer 62

GATE_PROJ

Top 16 Positive Activations
Rollout 8
$5$ of the cards in a row in a random order. She will be happy if and only
Rollout 4
^2}{(C_1 + C_2)^2} \quad \text{and}
Rollout 4
C_1 + C_2}\n\]\n\nGiven \( C_1 = 2000
Rollout 5
b = 175 \). Substituting these values:\n\[\nx = \sqrt{\
Rollout 5
{75^2 + 175^2}{2}} = \sqrt{\frac{
Rollout 8
. She shuffles the $10$ cards and lays out $5$ of the cards in a
Rollout 4
^2}{(C_1 + C_2)^2}\n\]\n\nSubstituting the values
Rollout 4
+ (0.16 \cdot 15)^2} = \sqrt{3.6
Rollout 8
order. She will be happy if and only if all the red cards laid out are adjacent and all the
Rollout 4
C_2)^2}\n\]\n\nSubstituting the values:\n\n\[\n\frac{\partial C
Rollout 8
will be happy if and only if all the red cards laid out are adjacent and all the green cards laid
Rollout 8
. She will be happy if and only if all the red cards laid out are adjacent and all the green
Rollout 5
}{3}\n\]\nSolving this, we find:\n\[\n3(b + 25)
Rollout 5
zoid into two regions of equal area. Find the greatest integer that does not exceed $x^2
Rollout 5
apezoid into two regions of equal area. Find the greatest integer that does not exceed $x^
Rollout 4
\text{pF} \), we first calculate the nominal equivalent capacitance:\n\n\[\nC_{
Top 16 Negative Activations
Rollout 2
: 9 *99 * (-1)^{997} mod125\n\nBecause
Rollout 2
, product becomes9*99*(-1)^{997} mod1000.
Rollout 6
$L$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work
Rollout 1
balls are not compact hence there is no guarantee.\n\nHences perhaps requires another approach, using the structure of
Rollout 1
= d sqrt(2} ~1.414 d > distances between centers d, hence intersection
Rollout 7
. Average=1350/18=75. Correct.\n\nNon-promoted:6
Rollout 1
over the Hilbert space, so thinking as in R^n, for which we know that even for three
Rollout 7
4.\n\nOriginal promoted P=16, R=8.\n\nAfter increase promoted total:79*(
Rollout 1
sqrt(2) �� 1.414d >= d, so intersection is non-empty
Rollout 4
8 +477.8388≈14,805.838
Rollout 0
of 20! -1} =2^{8-1}=128.\n\n**Final
Rollout 2
product modulo8:1*3*(-1)^{997}=1*3*(-1)^
Rollout 4
206 -1194)/2≈6 pF. Wait, but when we did
Rollout 1
0.\n\nSo the point y must be equidistant from all points in S, with distance d
Rollout 0
ordered pairs (a,b) would be 2^8 = 256. However, because
Rollout 2
999-9*1000=9999-9000=

UP_PROJ

Top 16 Positive Activations
Rollout 1
:\n\n-2〈x. This action would be the same as above for linear equations, but we already
Rollout 1
formal differences. Imagine that the set S, translated by y, becomes orthonormal, rescaled.
Rollout 6
, which might not be helpful.\n\nAlternatively, recall that P_angel is equivalent to P/poly,
Rollout 6
polynomial-time machine with a sparse oracle. (This is the Meyer, I think, theorem.) So perhaps
Rollout 6
be arbitrary. It is known that P/poly contains undecidable languages (as the advice can be
Rollout 6
, you can have for each n, a set of strings that encode the advice α_n. But since
Rollout 6
a polynomial-length advice string, which can be arbitrary. It is known that P/poly contains undec
Rollout 7
, after the error, all scores are increased by 5, so effectively, the new scores are original
Rollout 6
and only if it is polynomial-time Turing reducible to a sparse set. So that's the main
Rollout 6
the advice strings being generated non-uniformly. The class P/poly is equivalent to languages decidable
Rollout 1
Alternatively, another route:\n\nSince H is infinite-dimensional, embed the problem into a real Hilbert space,
Rollout 1
family can exist.\n\nMoreover, in Hilbert spaces, given any collection of vectors in a separated system like
Rollout 6
a polynomial-time machine \( M \). By the Meyer's theorem equivalence between \( \textbf{P
Rollout 6
is a p(n)-bit string, split it into p(n)/k chunks, each of k bits
Rollout 3
are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in the heavy gas radon
Rollout 8
*4*3*2)*(5)=2*120*5=1200
Top 16 Negative Activations
Rollout 2
by $1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem
Rollout 0
{}!$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure
Rollout 6
|} can be computed in poly-time given |x|, which is the length of x, then
Rollout 7
N/3\n\nTherefore:\n\nx(A -54) =56N/3 -18
Rollout 7
A +5)*x] / (P +x)=79.\n\nSimilarly, total non-promoted
Rollout 9
-5)/(10 - (10 - h)) = (-5)/h.\n\nThe equation is
Rollout 7
A +5)*x) / (P +x) )=75\n\nSimilarly, the new
Rollout 6
each m, there may be at most one n s.t. p(n) = m. Hence,
Rollout 7
Therefore:\n\n(N/12)(A -54) = 2N/3\n\nDivide
Rollout 0
be? Let me check with a smaller factorial.\n\nSuppose instead of 20!, take a simpler
Rollout 6
poly-time, the total time is polynomial in |x|. Therefore, L is in P, so P
Rollout 2
let's check with smaller products. For verification.\n\nSuppose instead of going up to 999
Rollout 7
0N => 3P=2N => P = (2/3)N.\n\nSo N
Rollout 7
N/3.\n\nTherefore,\n\nx(A -54)=2N/3.\n\nBut A is the
Rollout 1
⟩ - d2 / 2.\n\nSo, rearranged:\n\n⟨x, y⟩ = (⟨
Rollout 8
. Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We

DOWN_PROJ

Top 16 Positive Activations
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 1
{}y): \ x\in S\right\}\] \nis an orthonormal system of vectors
Rollout 6
p(n)$ , where $S^{=n} \subseteq S$ contains all the strings
Rollout 8
)! (choosing 5−r greens and permuting them). Thus, for a particular r
Rollout 2
10^{3}*10^{k-3}≡0*10^{k-3
Rollout 7
/3) +75x +59*(N/3) -59x =7
Rollout 6
a polynomial $p : \mathbb{N} \mapsto \mathbb{N}$
Top 16 Negative Activations
Rollout 1
why in the previous example with translated set S', it encountered a problem?\n\nAh, because in that example
Rollout 0
, 2^(3 -1)=4, which matches what we saw for 720.\n\n
Rollout 4
% error. \n\nBut since combining them by quadrature gives sqrt(0.52 +0.
Rollout 1
for a set S which includes 0, then it would require that 〈x, x'〉
Rollout 0
that the original rational number between 0 and 1 is written in reduced form, and you multiply numerator
Rollout 8
R1 G1) has both red and green cards, each as singleton blocks. Are singleton blocks considered
Rollout 1
0 and other points. Then, the act of translation introduced coordinates that conflicts with the required inner products.
Rollout 8
the colors. That is, the actual arrangement of colors in the sequence. So if in the sequence,
Rollout 8
in this case would consist of non-adjacent cards in color, but no, with two cards.\n\n
Rollout 8
by 3 greens. Wait, 2 reds but arranged as R, R in positions 1
Rollout 1
2/2)} / 2.\n\nBut perhaps it's hard.\n\nWait, back to the equation:\n\n
Rollout 3
required, but the problem says "compute", so it's expected to compute.\n\nAlternatively, perhaps check if
Rollout 3
70 m/s. But given compute precisely, it is about 168.5 m/s
Rollout 0
20! and a < b.\n\nThus previous logic applies; each such pair is obtained by assignment of
Rollout 0
written in reduced form, and you multiply numerator and denominator (from reduced form), and that product is equal
Rollout 0
are reduced to lowest terms. Which in original fraction it's not required that the numerator and denominator multiply to

Layer 63

GATE_PROJ

Top 16 Positive Activations
Rollout 8
be adjacent). But the way the problem is phrased: "all the red cards laid out are
Rollout 8
k) = 5! / (5 - k)!.\n\nSimilarly, arrange (5 -k)
Rollout 0
. So the count is half the number of coprime ordered pairs, or is it?\n\nWait,
Rollout 0
and had three prime factors. The number of coprime pairs a < b where a*b=7
Rollout 0
\neq b\). Hence, total coprime pairs are \(2^{8}\), where
Rollout 5
connecting the midpoints of the legs divides the trapezoid into two regions with areas in the ratio
Rollout 0
, but when we write a and b as coprime factors, the way I'm thinking is that
Rollout 0
(2,3,5), number of coprime pairs where a < b is 4 (
Rollout 3
πM)), whereas the root mean square speed is sqrt(3RT/M). Since the question refers to
Rollout 5
is parallel to the bases and that divides the trapezoid into two regions of equal area. Find
Rollout 0
assignments will satisfy a < b, even for coprime divisors. Wait, but in reality,
Rollout 7
$75$ and that of the non-promoted $59$ .\n(a) Find
Rollout 2
be ��0 mod8. Therefore, m=8n for some integer n.\n\nTherefore, x
Rollout 5
these are not similar necessarily. \n\nAlternatively, think algebraically. Let me denote the original trapez
Rollout 5
25) = 125y\n\nExpanding the left-hand side:\n\ny^2 -
Rollout 2
x≡109 mod1000.\n\nWait, but hold on, earlier we got x
Top 16 Negative Activations
Rollout 3
\) is the molar mass of radon in kg/mol.\n\n**Steps:**\n\n1. **
Rollout 8
athy has $5$ red cards and $5$ green cards. She shuffles the $10
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
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start by determining the equivalent capacitance \( C_{\text{eq}} \) and its uncertainty.
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>

UP_PROJ

Top 16 Positive Activations
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k) = 5! / (5 - k)!.\n\nSimilarly, arrange (5 -k)
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(n,k) = n! / (n - k)!.\n\nSimilarly, for a different color sequence with
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.327 / 1200 �� 0.003606
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6666... /1200 �� 0.003605
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0,150 /4975 �� Compute 4975*11
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C(5,5−r)*(5−r)! = 5!/( (5 - r
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000076996) �� 0.008775
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>=3, hence (10^k -1)≡-1 mod125.\n\nThus
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(5,5−r) *(5−r)! (choosing 5−r greens and
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76. sqrt(33,476)=183 m/s. But since the
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= 10! / (10-5)! = 10 * 9 *
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}=9*99*(-1). Because (-1)^997= -1, as9
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. So sqrt(13e-6) �� 0.003605
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mod8, so (10^k -1)≡(2^k -1). For
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.00007696) �� 0.008774
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(2^k -1) for k=1 to 999 mod 8.\n\nHmm
Top 16 Negative Activations
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oracle can contain those positions. So, for each (n, i), where i is a bit position
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.\n\nAnother angle: Let S_L contain the tuple <n, α_n >. But encode this in
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ounding to Significant Figures:**\n Given the inputs (molar mass \(222 \, \
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the *angel string* is $\textbf{not}$ similar to a *witness*
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n, create entries in S_L with the form (n, i, b) where b is the
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build a sparse oracle S_L consisting of all tuples <n, i, b>, where the i-th
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the length $n$ . Note that the *angel string* is $\textbf{not
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can be generated in polynomial time given the input length (i.e., 1^n), then the advice
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a unique identifier, perhaps the query is a pair (n, i) where i is a bit position
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counts, and divide by the total number of permutations (10P5).\n\nLet me try this approach
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**\n\n1. **Molar Mass of Radon (Rn):**\n Radon has a m
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molar mass of radon in kg/mol.\n\n**Steps:**\n\n1. **Molar Mass of
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2, excluding the case \(a = b\) (which is impossible since 20! is not
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isosceles with \(GE = GM\).\n\n4. **Equations of Lines**:\n
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the molar mass of radon in kg/mol.\n\n**Steps:**\n\n1. **Molar Mass
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to be the *angel string*for all $x$ of the length $n$

DOWN_PROJ

Top 16 Positive Activations
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$ units. Isosceles triangle $GEM$ has base $EM$ , and the
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', ⟨√2/d (x - y), ��2/d (x' - y)⟩
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$x \in \{0,1\}^n$ $$ x \in L
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.\n\nWe need to find y such that y ∈ ��_{x ∈ S} sphere(x, d
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each $n \in \mathbb{N}$ , the number of strings of length $
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pairwise distance d apart. To find y such that ��2/d (x - y) x ∈
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length $n \in \mathbb{N}$ . In other words, there is a poly
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{}y): \ x\in S\right\}\] \nis an orthonormal system of vectors
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of the altitude to $EM$ in $\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay
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Wait, wait subtract the lower from upper:\n\n= ��[0 to10] [ (5
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p(n)$ , where $S^{=n} \subseteq S$ contains all the strings
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a polynomial $p : \mathbb{N} \mapsto \mathbb{N}$
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Given $k \in \mathbb{N}$ sparse sets $S_1,
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the reds are not all together. Similarly, GGRGR would also not work. So we need
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. For example, card orders RRGGG, GGGGR, or RRRRR will make Kathy
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TM M, when processing x ∈ {0,1}^n, uses the advice string α_n
Top 16 Negative Activations
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) / 2 = 0.\n\nRearranging:\n\n||y||2 - ⟨x +
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-promoted $47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to
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but depending on which primes. But here is the crux: instead of trying to group, this is
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\mathbb{N} \mapsto \mathbb{N}$ such that for each
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a )2 + ( (a2)/(a + b)^2 * Δb )2 )\n\nLet
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0.02 \mathrm{~V}$. What is the percentage error in the calculation of
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a point $ y\in\mathcal{H}$ such that \n\[ \left\
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^2/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I
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6 = 30240.\n\nNumerator:\n\nFirst, Case 1: 2
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The length \( x \) is given by:\n\[\nx = \sqrt{\frac{a^
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stored in this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find
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be adjacent). But the way the problem is phrased: "all the red cards laid out are
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
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||2 = d2/2.\n\nRearranged:\n\n2⟨xi, y⟩ = ||
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>